d and f Block Elements
$\mathrm{NaCl}$ reacts with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ to give reddish fumes (B), which react with $\mathrm{NaOH}$ to give yellow solution (C). (B) and (C) respectively are ;
The electronic configuration for Neodymium is:
[Atomic Number for Neodymium 60]
A transition metal '$\mathrm{M}$' among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Fe}$ has the highest second ionisation enthalpy. The spin-only magnetic moment value of $\mathrm{M}^{+}$ ion is _______ BM (Near integer)
(Given atomic number $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$)
Explanation:
Identify which metal (M) has the highest second ionization enthalpy.
We are comparing the elements Sc, Ti, V, Cr, Mn, and Fe in terms of their second ionization enthalpy (IE₂). Recall:
The second ionization enthalpy (IE₂) is the energy required to remove one electron from the singly charged ion $ \mathrm{M}^{+} $ to form $ \mathrm{M}^{2+} $.
For most 3d transition metals, the first electron is removed from the 4s orbital.
Particularly large values of IE₂ often occur if, after removal of the first electron, one is forced to remove an electron from a stable configuration (e.g., half-filled d-orbital).
Let us outline the ground-state (neutral atom) electron configurations and the removal of the first and second electrons:
Sc $\bigl[ \mathrm{Ar} \bigr] 3d^1\,4s^2 $
$\mathrm{Sc} \rightarrow \mathrm{Sc}^{+}$: remove 1 electron from 4s
$\mathrm{Sc}^{+} \rightarrow \mathrm{Sc}^{2+}$: remove the second 4s electron
Final: $\mathrm{Sc}^{2+} = [\mathrm{Ar}]\,3d^1$
Ti $\bigl[ \mathrm{Ar} \bigr] 3d^2\,4s^2 $
1st electron from 4s → $\mathrm{Ti}^{+} = [\mathrm{Ar}]\,3d^2\,4s^1$
2nd electron from 4s → $\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$
V $\bigl[ \mathrm{Ar} \bigr] 3d^3\,4s^2 $
1st electron from 4s → $\mathrm{V}^{+} = [\mathrm{Ar}]\,3d^3\,4s^1$
2nd electron from 4s → $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$
Cr $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^1 $
1st electron from 4s → $\mathrm{Cr}^{+} = [\mathrm{Ar}]\,3d^5$ (a stable half-filled d$^{5}$ configuration)
2nd electron now must come from the 3d shell → $\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$
Because removing an electron from a half-filled d$^{5}$ orbital costs a lot of energy, $\mathrm{Cr}$ generally has a notably high second IE.
Mn $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^2 $
1st electron from 4s → $\mathrm{Mn}^{+} = [\mathrm{Ar}]\,3d^5\,4s^1$
2nd electron from 4s → $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ (thus still d$^{5}$ in the end)
The second ionization for Mn is large but less dramatic than for Cr because Mn$^{2+}$ ends with a half-filled d$^{5}$. You are not removing from a half-filled d$^{5}$ to get Mn$^{2+}$.
Fe $\bigl[ \mathrm{Ar} \bigr] 3d^6\,4s^2 $
1st electron from 4s → $\mathrm{Fe}^{+} = [\mathrm{Ar}]\,3d^6\,4s^1$
2nd electron from 4s → $\mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6$
From experimental data (and the arguments above), $\mathrm{Cr}$ indeed has the highest second ionization enthalpy among these six metals.
Determine the electronic configuration of $\mathbf{Cr}^{+}$ and its spin-only magnetic moment.
Neutral $\mathrm{Cr}$: $\bigl[ \mathrm{Ar} \bigr]\,3d^5\,4s^1$
$\mathrm{Cr}^{+}$: Removal of the 4s electron ⇒ $\bigl[ \mathrm{Ar} \bigr]\,3d^5$
The $3d^5$ configuration has 5 unpaired electrons.
The spin-only magnetic moment $\mu$ is given by
$ \mu = \sqrt{n(n+2)} \; \mathrm{BM}, $
where $n$ = number of unpaired electrons. Here $n = 5$, so
$ \mu = \sqrt{5(5+2)} \;=\; \sqrt{35} \;\approx\; 5.92 \;\text{BM}. $
This is often rounded to about 5.9 or 6.0 BM.
Number of colourless lanthanoid ions among the following is __________.
$\mathrm{Eu}^{3+}, \mathrm{Lu}^{3+}, \mathrm{Nd}^{3+}, \mathrm{La}^{3+}, \mathrm{Sm}^{3+}$
Explanation:
The color of lanthanoid ions in solutions is mainly due to the electronic transitions within the 4f subshell. The lanthanoid ions are more likely to be colorless when they have fully filled (with 14 electrons) or completely empty (with 0 electrons) f orbitals because, in these cases, there are no electrons to undergo f-f transitions, and as a result, no absorption of visible light occurs leading to colorlessness.
Let's consider the electronic configurations of the lanthanoid ions provided:
- $\mathrm{Eu}^{3+}$: Europium (Eu) has an atomic number of 63. Neutral europium ([Xe]4f7 6s^2) loses three electrons to form Eu3+, leaving it with an electronic configuration equivalent to [Xe]4f6. With 6 electrons in the f orbital, it can undergo f-f transitions, thus it is not colorless.
- $\mathrm{Lu}^{3+}$: Lutetium (Lu) has an atomic number of 71. In its 3+ ionic state, lutetium has lost its 6s and 5d electrons and is left with a completely filled 4f orbital ([Xe] 4f14). This configuration cannot allow for any f-f transitions, as there are no available energy levels within the f orbital for an electron to jump to, making Lu3+ colorless.
- $\mathrm{Nd}^{3+}$: Neodymium (Nd) has an atomic number of 60. In its 3+ state ([Xe] 4f3), it clearly has partially filled f orbitals, which can absorb visible light for f-f transitions, so it is not colorless.
- $\mathrm{La}^{3+}$: Lanthanum (La) has an atomic number of 57. In its 3+ ionic state, it has a configuration of [Xe], meaning that its 4f orbital is completely empty. Since there are no electrons in the f orbital to undergo f-f transitions, La3+ is colorless.
- $\mathrm{Sm}^{3+}$: Samarium (Sm) has an atomic number of 62. As a 3+ ion ([Xe] 4f5), it too has electrons in the f orbital capable of undergoing f-f transitions, so it is not colorless.
From the analysis, the colorless lanthanoid ions among the ones listed are $\mathrm{Lu}^{3+}$ and $\mathrm{La}^{3+}$.
Therefore, the number of colorless lanthanoid ions among the given options is 2.
Among $\mathrm{VO}_2^{+}, \mathrm{MnO}_4^{-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, the spin-only magnetic moment value of the species with least oxidising ability is __________ BM (Nearest integer).
(Given atomic member $\mathrm{V}=23, \mathrm{Mn}=25, \mathrm{Cr}=24$)
Explanation:
In order to determine the spin-only magnetic moment value of the species with the least oxidizing ability, we first need to analyze their oxidation states and electron configurations.
1. $\mathrm{VO}_2^{+}$:
For vanadium in $\mathrm{VO}_2^{+}$, the oxidation state is +5. The electronic configuration of V is $[Ar] \, 3d^3 \, 4s^2$. Thus, in +5 oxidation state, vanadium will have zero d-electrons (since it loses 5 electrons) and, consequently, $\mathrm{VO}_2^{+}$ is diamagnetic (since zero unpaired electrons).
2. $\mathrm{MnO}_4^{-}$:
For manganese in $\mathrm{MnO}_4^{-}$, the oxidation state is +7. The electronic configuration of Mn is $[Ar] \, 3d^5 \, 4s^2$. Thus, in +7 oxidation state, manganese will have zero d-electrons (since it loses 7 electrons) and, consequently, $\mathrm{MnO}_4^{-}$ is also diamagnetic (since zero unpaired electrons).
3. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$:
For chromium in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, we have two chromium atoms. Each chromium is in the +6 oxidation state. The electronic configuration of Cr is $[Ar] \, 3d^5 \, 4s^1$. Thus, in +6 oxidation state, each chromium will have zero d-electrons (since it loses 6 electrons) and, consequently, $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also diamagnetic (since zero unpaired electrons).
From these observations, we note that all the species we analyzed are diamagnetic. However, to identify the species with the least oxidizing ability, we look at their standard reduction potentials (E° values).
Typically, the oxidizing ability increases with increasing positive E° values. Given the nature of these species, we can infer that:
- $\mathrm{MnO}_4^{-}$ is a very strong oxidizing agent (very high E° value).
- $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also a strong oxidizing agent (high but less than $\mathrm{MnO}_4^{-}$).
- $\mathrm{VO}_2^{+}$ has a moderate oxidizing ability (lesser E° value than both $\mathrm{MnO}_4^{-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$).
Therefore, among the given species, $\mathrm{VO}_2^{+}$ has the least oxidizing ability.
As mentioned previously, $\mathrm{VO}_2^{+}$ has zero unpaired electrons, making it diamagnetic. The spin-only magnetic moment is given by the formula:
$\mu = \sqrt{n(n+2)}$
where $n$ is the number of unpaired electrons.
For $\mathrm{VO}_2^{+}$, $n=0$, thus:
$\mu = \sqrt{0(0+2)} = 0 \ \text{BM}$
Given the options, the nearest integer value of the spin-only magnetic moment for the species with the least oxidizing ability (which is $\mathrm{VO}_2^{+}$) is indeed:
0 BM
Among $\mathrm{CrO}, \mathrm{Cr}_2 \mathrm{O}_3$ and $\mathrm{CrO}_3$, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _________ $10^{-2} \mathrm{BM}$ (nearest integer).
(Given atomic number of $\mathrm{Cr}$ is 24 )
Explanation:
First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.
1. $\mathrm{CrO}$: In $\mathrm{CrO}$, the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$. In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $3d^4$.
To find the spin-only magnetic moment, we use the formula:
$\mu = \sqrt{n(n+2)} \mathrm{BM}$
where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$, we have 4 unpaired electrons in the 3d orbitals.
Thus,
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mathrm{BM}$
2. $\mathrm{Cr}_2 \mathrm{O}_3$: In $\mathrm{Cr}_2 \mathrm{O}_3$, the oxidation state of chromium is +3. The electronic configuration of $\mathrm{Cr}^{3+}$ is $3d^3$ after losing three electrons.
For $\mathrm{Cr}^{3+}$, there are 3 unpaired electrons.
Thus,
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$
3. $\mathrm{CrO}_3$: In $\mathrm{CrO}_3$, the oxidation state of chromium is +6. The electronic configuration of $\mathrm{Cr}^{6+}$ is $3d^0$ after losing six electrons. There are no unpaired electrons for $\mathrm{Cr}^{6+}$.
Since $\mathrm{Cr}^{6+}$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.
The magnetic moment values for each compound are:
- $\mathrm{CrO}$: 4.90 BM (basic oxide)
- $\mathrm{Cr}_2 \mathrm{O}_3$: 3.87 BM (amphoteric oxide)
- $\mathrm{CrO}_3$: 0 BM
The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:
$4.90 + 3.87 = 8.77 \mathrm{BM}$
Expressing in terms of $10^{-2} \mathrm{BM}$,
$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$
Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $\times 10^{-2} \mathrm{BM}$ (nearest integer).
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products $\mathrm{A}$ and $\mathrm{B}$ along with the evolution of $\mathrm{CO}_2$. The sum of spin-only magnetic moment values of A and B is _________ B.M. (Nearest integer)
[Given atomic number : $\mathrm{C}: 6, \mathrm{Na}: 11, \mathrm{O}: 8, \mathrm{Fe}: 26, \mathrm{Cr}: 24$]
Explanation:
To determine the spin-only magnetic moments of products A and B formed from the fusion of chromite ore with sodium carbonate in the presence of air, we first need to examine the given reaction and the properties of the products.
The reaction provided is:
$4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 (\mathrm{A}) + 2 \mathrm{Fe}_2 \mathrm{O}_3 (\mathrm{B}) + 8 \mathrm{CO}_2$
Identifying the Products:- Product A is sodium chromate, $\mathrm{Na}_2 \mathrm{CrO}_4$.
- Product B is ferric oxide, $\mathrm{Fe}_2 \mathrm{O}_3$.
Determining the Magnetic Moments:
- For Product A ($\mathrm{Na}_2 \mathrm{CrO}_4$):
- In $\mathrm{Na}_2 \mathrm{CrO}_4$, the chromium is in the +6 oxidation state.
- The electronic configuration of $\mathrm{Cr}^{6+}$ is $[\mathrm{Ar}] 3d^0$.
- Since there are no unpaired electrons in the $3d$ orbitals, the spin-only magnetic moment is zero.
$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{0(0+2)} \text{ BM} = 0 \text{ BM} $
- For Product B ($\mathrm{Fe}_2 \mathrm{O}_3$):
- In $\mathrm{Fe}_2 \mathrm{O}_3$, the iron is in the +3 oxidation state.
- The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$.
- $\mathrm{Fe}^{3+}$ has 5 unpaired electrons in the $3d$ orbitals.
- The spin-only magnetic moment can be calculated using the formula:
$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{5(5+2)} \text{ BM} = \sqrt{35} \text{ BM} \approx 5.9 \text{ BM} $
Sum of Magnetic Moments:- The spin-only magnetic moment of Product A is $0 \text{ BM}$.
- The spin-only magnetic moment of Product B is $5.9 \text{ BM}$.
Therefore, the sum of the spin-only magnetic moments of A and B is:
$ 0 + 5.9 \approx 6 \text{ BM}$
Conclusion:The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).
In a borax bead test under hot condition, a metal salt (one from the given) is heated at point B of the flame, resulted in green colour salt bead. The spin-only magnetic moment value of the salt is _______ BM (Nearest integer) [Given atomic number of $\mathrm{Cu}=29, \mathrm{Ni}=28, \mathrm{Mn}=25, \mathrm{Fe}=26$]
Explanation:
Green coloured salt bead represents the metal ion taken is $\mathrm{Fe}^{3+}$ so, $\mathrm{Fe}^{3+}:[\mathrm{Ar}] 4 \mathrm{~s}^0 3 d^5$
So, $\mu=\sqrt{5 \times 7}=5.9=6$
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula $\mathrm{M}_2 \mathrm{O}_{\mathrm{n}}$ (where $\mathrm{n}=3,4,5$). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is _________ $\mathrm{BM}$ (near integer)
(Given atomic number: $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26, \mathrm{Co}: 27, \mathrm{Ni}: 28, \mathrm{Cu}: 29, \mathrm{Zn}: 30$)
Explanation:
Vanadium has highest enthalpy of atomization among first row transition elements.
$\mathrm{V}_2 \mathrm{O}_5$ is amphoteric
In $\mathrm{V}^{5+}$ there are no unpaired electrons.
Thus, $\mu=0$
Consider the following reaction
$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O} \text {. }$
Product '$\mathrm{A}$' in neutral or acidic medium disproportionate to give products '$\mathrm{B}$' and '$\mathrm{C}$' along with water. The sum of spin-only magnetic moment values of $\mathrm{B}$ and $\mathrm{C}$ is ________ BM. (nearest integer) (Given atomic number of $\mathrm{Mn}$ is 25)
Explanation:
$\mathrm{A}$ is $\mathrm{K}_2 \mathrm{MnO}_4$
$\mathrm{B}$ and $\mathrm{C}$ are $\mathrm{KMnO}_4$ and $\mathrm{MnO}_2$
$\mathrm{KMnO}_4(\mu=0)$
$\mathrm{MnO}_2(\mathrm{Mn}^{4+})(\mu=3.87)$
Sum $=3.87=4$ (Nearest integer)
Total number of ions from the following with noble gas configuration is _________.
$\mathrm{Sr}^{2+}(z=38), \mathrm{Cs}^{+}(z=55), \mathrm{La}^{2+}(z=57), \mathrm{Pb}^{2+}(z=82), \mathrm{Yb}^{2+}(z=70)$ and $\mathrm{Fe}^{2+}(z=26)$
Explanation:
To determine which of these ions have a noble gas configuration, we must consider what the phrase "noble gas configuration" means. Atoms or ions with a noble gas configuration have completely filled electron shells, similar to the electron configuration of the noble gases, which are the elements found in Group 18 of the periodic table. Noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).
Let's examine each ion given:
- $\mathrm{Sr}^{2+}(z=38)$: Strontium has atomic number 38 and loses 2 electrons to form a $\mathrm{Sr}^{2+}$ ion, giving it the same electron configuration as krypton (Kr), with atomic number 36. Hence, it has a noble gas configuration.
- $\mathrm{Cs}^{+}(z=55)$: Cesium has atomic number 55, and by losing 1 electron to form $\mathrm{Cs}^{+}$, it has the electron configuration of xenon (Xe), with atomic number 54. Thus, it also has a noble gas configuration.
- $\mathrm{La}^{2+}(z=57)$: Lanthanum has atomic number 57. If it lost 2 electrons to form $\mathrm{La}^{2+}$, it would not have a noble gas configuration because it would have one electron more than xenon (Xe), which means it would not fully match any noble gas electron configuration.
- $\mathrm{Pb}^{2+}(z=82)$: Lead has an atomic number of 82, so when it loses 2 electrons to form $\mathrm{Pb}^{2+}$, it has 80 electrons, which is the same electron number as mercury (Hg) and not a noble gas. Therefore, $\mathrm{Pb}^{2+}$ does not have a noble gas configuration.
- $\mathrm{Yb}^{2+}(z=70)$: Ytterbium has an atomic number of 70. By losing 2 electrons to form $\mathrm{Yb}^{2+}$, it has 68 electrons, aligning with the electron configuration of Erbium (Er). $\mathrm{Yb}^{2+}$ and not a noble gas.
- $\mathrm{Fe}^{2+}(z=26)$: Iron has an atomic number of 26. When it becomes $\mathrm{Fe}^{2+}$ by losing 2 electrons, it has 24 electrons. This does not correspond to any noble gas configuration, as the nearest noble gas is argon (Ar) with 18 electrons.
${\left[\mathrm{Sr}^{2+}\right]=[\mathrm{Kr}]}$
${\left[\mathrm{Cs}^{+}\right]=[\mathrm{Xe}]}$
${\left[\mathrm{La}^{2+}\right]=[\mathrm{Xe}] 5 \mathrm{~d}^1}$
${\left[\mathrm{~Pb}^{2+}\right]=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^2}$
${\left[\mathrm{Yb}^{2+}\right]=[Xe]4f^{14}}$
${\left[\mathrm{Fe}^{2+}\right]=[\mathrm{Ar}] 3 \mathrm{~d}^6}$
Therefore, the ions $\mathrm{Sr}^{2+}$ and $\mathrm{Cs}^{+}$ have a noble gas configuration. Altogether, there are 2 ions with a noble gas configuration.
The pair of lanthanides in which both elements have high third - ionization energy is :
Given below are two statement: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A: $5 \mathrm{f}$ electrons can participate in bonding to a far greater extent than $4 \mathrm{f}$ electrons
Reason R: $5 \mathrm{f}$ orbitals are not as buried as $4 \mathrm{f}$ orbitals
In the light of the above statements, choose the correct answer from the options given below
Prolonged heating is avoided during the preparation of ferrous ammonium sulphate to :
Which of the following statements are correct?
(A) The M$^{3+}$/M$^{2+}$ reduction potential for iron is greater than manganese.
(B) The higher oxidation states of first row d-block elements get stabilized by oxide ion.
(C) Aqueous solution of Cr$^{2+}$ can liberate hydrogen from dilute acid.
(D) Magnetic moment of V$^{2+}$ is observed between 4.4 - 5.2 BM.
Choose the correct answer from the options given below :
Which halogen is known to cause the reaction given below :
$2 \mathrm{Cu}^{2+}+4 \mathrm{X}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{X}_{2}(\mathrm{s})+\mathrm{X}_{2}$
In chromyl chloride, the number of d-electrons present on chromium is same as in
(Given at no. of $\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$ )
Element not present in Nessler's reagent is :
Given below are two statements: one is labelled as "Assertion A" and the other is labelled as "Reason R"
Assertion A : In the complex $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\mathrm{Fe}(\mathrm{CO})_{5}$, the metals have zero oxidation state.
Reason R : Low oxidation states are found when a complex has ligands capable of $\pi$-donor character in addition to the $\sigma$-bonding.
In the light of the above statements, choose the most appropriate answer from the options given below
During the reaction of permanganate with thiosulphate, the change in oxidation of manganese occurs by value of 3. Identify which of the below medium will favour the reaction.
Strong reducing and oxidizing agents among the following, respectively, are :
Which element is not present in Nessler's reagent?
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : $\mathrm{Cu^{2+}}$ in water is more stable than $\mathrm{Cu^{+}}$.
Reason (R) : Enthalpy of hydration for $\mathrm{Cu^{2+}}$ is much less than that of $\mathrm{Cu^{+}}$.
In the light of the above statements, choose the correct answer from the options given below :
Highest oxidation state of Mn is exhibited in $\mathrm{Mn_2O_7}$. The correct statements about $\mathrm{Mn_2O_7}$ are
(A) Mn is tetrahedrally surrounded by oxygen atoms.
(B) Mn is octahedrally surrounded by oxygen atoms.
(C) Contains Mn-O-Mn bridge.
(D) Contains Mn-Mn bond.
Choose the correct answer from the options given below :
When $\mathrm{Cu}^{2+}$ ion is treated with $\mathrm{KI}$, a white precipitate, $\mathrm{X}$ appears in solution. The solution is titrated with sodium thiosulphate, the compound $\mathrm{Y}$ is formed. $\mathrm{X}$ and $\mathrm{Y}$ respectively are :
| $\mathrm{X=CuI_2}$ | $\mathrm{Y=Na_2S_4O_6}$ |
| $\mathrm{X=Cu_2I_2}$ | $\mathrm{Y=Na_2S_4O_6}$ |
| $\mathrm{X=CuI_2}$ | $\mathrm{Y=Na_2S_2O_3}$ |
| $\mathrm{X=Cu_2I_2}$ | $\mathrm{Y=Na_2S_4O_5}$ |
$\mathrm{Nd^{2+}}$ = __________
The correct order of basicity of oxides of vanadium is :
During the qualitative analysis of $\mathrm{SO}_{3}^{2-}$ using dilute $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}$ gas is evolved which turns $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution (acidified with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$) :
Given below are two statements :
Statement I : Nickel is being used as the catalyst for producing syn gas and edible fats.
Statement II : Silicon forms both electron rich and electron deficient hydrides.
In the light of the above statements, choose the most appropriate answer from the options given below :
The set of correct statements is :
(i) Manganese exhibits +7 oxidation state in its oxide.
(ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.
(iii) Sc shows +4 oxidation state which is oxidizing in nature.
(iv) Cr shows oxidising nature in +6 oxidation state.
An indicator 'X' is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with $\mathrm{H_2O_2}$ at room temp. The indicator 'X' forms blue coloured complex with compound 'A' present in the solution. The indicator 'X' and compound 'A' respectively are :
A chloride salt solution acidified with dil.HNO$_3$ gives a curdy white precipitate, [A], on addition of AgNO$_3$. [A] on treatment with NH$_4$OH gives a clear solution B. A and B are respectively :
Which one amongst the following are good oxidizing agents?
A. Sm$^{2+}$
B. Ce$^{2+}$
C. Ce$^{4+}$
D. Tb$^{4+}$
Choose the most appropriate answer from the options given below :
An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is :
Explanation:
The oxidation state of oxygen is usually -2 and the oxidation state of chlorine is usually -1.
Therefore, we can write:
$\mathrm{Cr} + 2\mathrm{O}(-2) + 2\mathrm{Cl}(-1) = 0$
Simplifying the equation, we get:
$\mathrm{Cr} - 4 - 2 = 0$
$\mathrm{Cr} - 6 = 0$
$\mathrm{Cr} = +6$
Therefore, the oxidation state of chromium in chromyl chloride is $(+6)$.
Explanation:
$\mathrm{MnO_4^-} + \mathrm{I^-} + \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + \mathrm{I_2} + \mathrm{H_2O}$
In this reaction, $\mathrm{KMnO}_4$ acts as an oxidizing agent, while $\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\mathrm{KMnO_4}$ to +2 in $\mathrm{Mn^{2+}}$, which is a reduction of 5 units.
Therefore, the total change in the oxidation state of manganese involved in the reaction is $\boxed{5}$.
Among following compounds, the number of those present in copper matte is ___________.
A. $\mathrm{CuCO_{3}}$
B. $\mathrm{Cu_{2}S}$
C. $\mathrm{Cu_{2}O}$
D. $\mathrm{FeO}$
Explanation:
Copper matte is a mixture of copper sulfide and iron sulfide formed as a result of smelting of copper ore. It contains mainly copper sulfide (Cu2S) as the major component and iron sulfide (FeS) as a minor component.
The compounds A. CuCO3, C. Cu2O, and D. FeO are not present in copper matte, but can be formed as a result of further processing or refining of the matte.
Therefore, among the given compounds, only "B. Cu2S" is present in copper matte.
The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is _____________.
Explanation:
$\mathrm{3{e^ - } + 4{H^ + } + MnO_4^ - \buildrel {} \over \longrightarrow Mn{O_2} + 2{H_2}O}$
How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? ______________
(Given : Atomic number V, 23; Cr, 24; Fe, 26; Ni, 28)
V$^{3+}$, Cr$^{3+}$, Fe$^{2+}$, Ni$^{3+}$
Explanation:
$ \begin{array}{ll} & n \\ \mathrm{V}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^2 4 \mathrm{~s}^0 & 2 \\\\ \mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^0 & 3 \\\\ \mathrm{Fe}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0 & 4 \\\\ \mathrm{Ni}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^7 4 \mathrm{~s}^0 & 3 \end{array} $
$\mathrm{Cr}^{3+} ~\&~ \mathrm{Ni}^{3+}$ have same value of $\mu_{\mathrm{s}}$
Which of the following $3\mathrm{d}$-metal ion will give the lowest enthalpy of hydration $\left(\Delta_{\text {hyd }} \mathrm{H}\right)$ when dissolved in water ?
The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives :
In following pairs, the one in which both transition metal ions are colourless is :
In neutral or faintly alkaline medium, $\mathrm{KMnO}_{4}$ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :
Match List - I with List - II, match the gas evolved during each reaction.
| List - I | List - II | ||
|---|---|---|---|
| (A) | $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow}$ | (I) | $\mathrm{H}_{2}$ |
| (B) | $\mathrm{KMnO}_{4}+\mathrm{HCl} \rightarrow$ | (II) | $\mathrm{N}_{2}$ |
| (C) | $\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \rightarrow$ | (III) | $\mathrm{O}_{2}$ |
| (D) | $\mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow}$ | (IV) | $\mathrm{Cl}_{2}$ |
Choose the correct answer from the options given below :
Which of the following has least tendency to liberate $\mathrm{H}_{2}$ from mineral acids?
In neutral or alkaline solution, $\mathrm{MnO}_{4}^{-}$ oxidises thiosulphate to :
Given below are two statements.
Statement I : Iron (III) catalyst, acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and neutral $\mathrm{KMnO}_{4}$ have the ability to oxidise $\mathrm{I}^{-}$ to $\mathrm{I}_{2}$ independently.
Statement II : Manganate ion is paramagnetic in nature and involves $\mathrm{p} \pi-\mathrm{p} \pi$ bonding.
In the light of the above statements, choose the correct answer from the options given below.
The total number of $\mathrm{Mn}=\mathrm{O}$ bonds in $\mathrm{Mn}_{2} \mathrm{O}_{7}$ is __________.
