p-Block Elements
$\mathrm{Cl}_2 \mathrm{O}_7, \mathrm{CO}, \mathrm{PbO}_2, \mathrm{~N}_2 \mathrm{O}, \mathrm{NO}, \mathrm{Al}_2 \mathrm{O}_3, \mathrm{SiO}_2, \mathrm{~N}_2 \mathrm{O}_5, \mathrm{SnO}_2$
Explanation:
An amphoteric oxide is one that displays both acidic and basic properties; such substances react with both acids and bases. In the p-block, several elements can form amphoteric oxides, particularly those elements that are metallic or metalloid in nature. It's essential to recognize which oxides among the provided list have an amphoteric character:
- $\mathrm{Cl}_2 \mathrm{O}_7$ - This is dichlorine heptoxide, an oxide of chlorine, and shows acidic properties. It is not amphoteric.
- $\mathrm{CO}$ - Carbon monoxide does not behave as an acidic or basic oxide; it is mainly a neutral oxide.
- $\mathrm{PbO}_2$ - Lead(IV) oxide, also known as plumbic oxide. This oxide of lead is amphoteric, but it is much less common and less amphoteric than lead(II) oxide, PbO.
- $\mathrm{N}_2 \mathrm{O}$ - Nitrous oxide is a neutral oxide; it does not exhibit acidic or basic properties.
- $\mathrm{NO}$ - Nitric oxide, like N2O, is a neutral oxide.
- $\mathrm{Al}_2 \mathrm{O}_3$ - Aluminium oxide is well known for being amphoteric. It reacts with both acids and bases to produce salts and water.
- $\mathrm{SiO}_2$ - Silicon dioxide is generally acidic. Although it does not react with water to form an acidic solution, it will react with basic oxides. Therefore, it is typically not considered amphoteric.
- $\mathrm{N}_2 \mathrm{O}_5$ - Dinitrogen pentoxide is an acidic oxide; it is the anhydride of nitric acid.
- $\mathrm{SnO}_2$ - Tin dioxide, also known as stannic oxide. This oxide of tin is amphoteric, reacting with both acids and bases.
From the list, the amphoteric oxides are:
- $\mathrm{PbO}_2$ - Lead(IV) oxide (less commonly amphoteric compared to PbO)
- $\mathrm{Al}_2 \mathrm{O}_3$ - Aluminium oxide
- $\mathrm{SnO}_2$ - Tin dioxide
So, the number of amphoteric oxides in the list provided is 3.
1 mole of $\mathrm{PbS}$ is oxidised by "$\mathrm{X}$" moles of $\mathrm{O}_3$ to get "$\mathrm{Y}$" moles of $\mathrm{O}_2$. $\mathrm{X}+\mathrm{Y}=$ _________.
Explanation:
$\begin{aligned} & \mathrm{PbS}+4 \mathrm{O}_3 \rightarrow \mathrm{PbSO}_4+4 \mathrm{O}_2 \\ & \mathrm{x}=4, \mathrm{y}=4 \end{aligned}$
From the given list, the number of compounds with +4 oxidation state of Sulphur ________.
$\mathrm{SO}_3, \mathrm{H}_2 \mathrm{SO}_3, \mathrm{SOCl}_2, \mathrm{SF}_4, \mathrm{BaSO}_4, \mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7 $
Explanation:
| Compounds | $\mathrm{SO_3}$ | $\mathrm{H_2SO_3}$ | $\mathrm{SOCl_2}$ | $\mathrm{SF_4}$ | $\mathrm{BaSO_4}$ | $\mathrm{H_2S_2O_7}$ |
|---|---|---|---|---|---|---|
| O.S. of Sulphur: | +6 | +4 | +4 | +4 | +6 | +6 |
To determine the number of compounds with a +4 oxidation state of sulfur, we need to examine the oxidation states of sulfur in each of the listed compounds. The +4 oxidation state means that sulfur has lost 4 electrons compared to its elemental state.
Let's go through each compound:
- $\mathrm{SO}_3$: Sulfur trioxide. In this compound, sulfur exhibits a +6 oxidation state because each oxygen contributes -2, for a total of -6 (3 oxygens), which must be balanced by sulfur to maintain a neutral charge. So, this compound does not have sulfur in the +4 oxidation state.
- $\mathrm{H}_2 \mathrm{SO}_3$: Sulfurous acid. Here again, with two hydrogens (each contributing +1 = total +2) and three oxygens (each contributing -2 = total -6), sulfur has an oxidation state of +4 to balance out the -4 from the oxygens and +2 from the hydrogens. This compound does have sulfur in the +4 oxidation state.
- $\mathrm{SOCl}_2$: Thionyl chloride. Sulfur in this compound is connected to two chlorine atoms and one oxygen atom. Chlorine generally has an oxidation state of -1 (total -2 for both Cl atoms) and oxygen -2. To balance the -4 charge (from one oxygen and two chlorines), sulfur must have an oxidation state of +4. This compound has sulfur in the +4 oxidation state.
- $\mathrm{SF}_4$: Sulfur tetrafluoride. Fluorine is almost always in the -1 oxidation state, and there are four fluorine atoms for a total of -4. To balance this, sulfur must have a +4 oxidation state, making this compound one with sulfur in the +4 oxidation state.
- $\mathrm{BaSO}_4$: Barium sulfate. Barium has a +2 oxidation state, and sulfate (SO4) as a whole must have a -2 oxidation state to balance the barium. In sulfate, sulfur has an oxidation state of +6 (since each oxygen is -2 for a total of -8, and +6 from sulfur balances it to -2 overall). Hence, sulfur does not have a +4 oxidation state in this compound.
- $\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7$: Pyrosulfuric acid or oleum. Each hydrogen is +1 (total +2), and the seven oxygens are -2 each (total -14). To balance -12 (total of oxygens and hydrogens), each sulfur must have an oxidation state of +6. Therefore, sulfur is not in the +4 oxidation state in this molecule.
So, from the given list, only three compounds have sulfur in the +4 oxidation state: $\mathrm{H}_2 \mathrm{SO}_3$, $\mathrm{SOCl}_2$, and $\mathrm{SF}_4$. Therefore, the number of compounds with a +4 oxidation state of sulfur is three.
The correct group of halide ions which can be oxidised by oxygen in acidic medium is :
The covalency and oxidation state respectively of boron in $\left[\mathrm{BF}_{4}\right]^{-}$, are :
The incorrect statement from the following for borazine is :
Given below are two statements :
Statement I : Boron is extremely hard indicating its high lattice energy.
Statement II : Boron has highest melting and boiling point compared to its other group members.
In the light of the above statements, choose the most appropriate answer from the options given below :
Given below are two statements :
Statement I : $\mathrm{SbCl}_{5}$ is more covalent than $\mathrm{SbCl}_{3}$
Statement II: The higher oxides of halogens also tend to be more stable than the lower ones.
In the light of the above statements, choose the most appropriate answer from the options given below :
Which one of the following pairs is an example of polar molecular solids?
One mole of $\mathrm{P}_{4}$ reacts with 8 moles of $\mathrm{SOCl}_{2}$ to give 4 moles of $\mathrm{A}, x$ mole of $\mathrm{SO}_{2}$ and 2 moles of $\mathrm{B} . \mathrm{A}, \mathrm{B}$ and $x$ respectively are :
For compound having the formula $\mathrm{GaAlCl}_{4}$, the correct option from the following is :
Match List I with List II
| LIST I Name of reaction |
LIST II Reagent used |
||
|---|---|---|---|
| A. | Hell-Volhard-Zelinsky reaction | I. | $NaOH + {I_2}$ |
| B. | Iodoform reaction | II. | (i) $Cr{O_2}C{l_2},C{S_2}$ (ii) ${H_2}O$ |
| C. | Etard reaction | III. | (i) $B{r_2}$ / red phosphorus (ii) $H_2O$ |
| D. | Gatterman-Koch reaction | IV. | $CO,HCl,$, anhyd. $AlC{l_3}$ |
Choose the correct answer from the options given below:
The correct order of bond enthalpy ($\mathrm{kJ~mol^{-1}}$) is :
For electron gain enthalpies of the elements denoted as $\Delta_{\mathrm{eg}} \mathrm{H}$, the incorrect option is :
Given below are two statements:
Statement I : Chlorine can easily combine with oxygen to form oxides; and the product has a tendency to explode.
Statement II : Chemical reactivity of an element can be determined by its reaction with oxygen and halogens.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green
Statement II: The green colour observed is due to the formation of copper(I) metaborate
In light of the above statements, choose the most appropriate answer from the options given below :
During the borax bead test with $\mathrm{CuSO_4}$, a blue green colour of the bead was observed in oxidising flame due to the formation of :
"A" obtained by Ostwald's method involving air oxidation of NH$_3$, upon further air oxidation produces "B", "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test.
Identify A and B, respectively :
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Carbon forms two important oxides - CO and CO$_2$. CO is neutral whereas CO$_2$ is acidic in nature
Reason R : CO$_2$ can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water
In the light of the above statements, choose the most appropriate answer from the options given below :
A. Ammonium salts produce haze in atmosphere.
B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.
C. Polychlorinated biphenyls act as cleansing solvents.
D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water.
Choose the correct answer from the options given below :
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from :
Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively :
Compound A reacts with NH$_4$Cl and forms a compound B. Compound B reacts with H$_2$O and excess of CO$_2$ to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively :
$\mathrm{K_2Cr_2O_7}$ paper acidified with dilute $\mathrm{H_2SO_4}$ turns green when exposed to :
Which of the Phosphorus oxoacid can create silver mirror from $\mathrm{AgNO_3}$ solution?
If the formula of Borax is $\mathrm{Na_2B_4O_x(OH)_y~.~zH_2O}$, then $x+y+z=$ ___________
Explanation:
Comparing this to the given formula, we can assign the values for x, y, and z:
x = 5 (from O₅)
y = 4 (from (OH)₄)
z = 8 (from 8H₂O)
Now, let's sum up the values of x, y, and z:
$x+y+z=5+4+8=17$
So, the value of $x+y+z$ is indeed 17.
The difference in the oxidation state of $\mathrm{Xe}$ between the oxidised product of $\mathrm{Xe}$ formed on complete hydrolysis of $\mathrm{XeF}_{4}$ is ___________
Explanation:
in $\mathrm{XeO}_3$, Oxidation state of $\mathrm{Xe}=+6$
in $\mathrm{XeF}_4$, Oxidation state of $\mathrm{Xe}=+4$
So difference in oxidation state $=2$
In the following reactions, the total number of oxygen atoms in X and Y is ___________.
Na$_2$O + H$_2$O $\to$ 2X
Cl$_2$O$_7$ + H$_2$O $\to$ 2Y
Explanation:
$X$ has one $O$ and $Y$ has four $O$
The ratio of sigma and $\pi$ bonds present in pyrophosphoric acid is ___________.
Explanation:
$ \frac{\text { No. of } \sigma \text { bonds }}{\text { No. of } \pi \text { bonds }}=\frac{12}{2}=6 $
$\mathrm{XeF}_{4}$ reacts with $\mathrm{SbF}_{5}$ to form
$[\mathrm{XeF}_{m}]^{\mathrm{n}+}\left[\mathrm{SbF}_{y}\right]^{z-}$.
$\mathrm{m}+\mathrm{n}+\mathrm{y}+\mathrm{z}=$ __________
Explanation:
The reaction of $XeF_4$ with $SbF_5$ is a known reaction that forms a complex compound $XeF_3^+$ and $SbF_6^−$. Here, the $XeF_4$ is providing one fluoride ion to $SbF_5$, resulting in $XeF_3^+$ and $SbF_6^−$.
In the form $[XeF_m]^{n+}[SbF_y]^{z-}$, the parameters m, n, y, and z correspond to the following:
- m is the number of fluoride ions attached to Xe.
- n is the charge of the $XeF_m$ cation.
- y is the number of fluoride ions attached to Sb.
- z is the charge of the $SbF_y$ anion.
Given the formation of $XeF_3^+$ and $SbF_6^−$, the values would be as follows:
- m = 3 (from $XeF_3$)
- n = 1 (from $XeF_3^+$)
- y = 6 (from $SbF_6$)
- z = 1 (from $SbF_6^−$)
Therefore, $m + n + y + z = 3 + 1 + 6 + 1 = 11$.
Sum of $\pi$-bonds present in peroxodisulphuric acid and pyrosulphuric acid is ___________
Explanation:
Peroxodisulphuric acid -
No. of $\pi $ – bonds = 4
Pyro sulphuric acid -
No. of $\pi $ – bonds = 4
Total $\pi $ – bonds = 4 + 4 = 8
In liquation process used for tin (Sn), the metal :
Given below are two statements.
Statement I : Stannane is an example of a molecular hydride.
Statement II : Stannane is a planar molecule.
In the light of the above statement, choose the most appropriate answer from the options given below.
When borax is heated with $\mathrm{CoO}$ on a platinum loop, blue coloured bead formed is largely due to :
Dinitrogen is a robust compound, but reacts at high altitudes to form oxides. The oxide of nitrogen that can damage plant leaves and retard photosynthesis is :
White phosphorus reacts with thionyl chloride to give :
Concentrated $\mathrm{HNO}_{3}$ reacts with Iodine to give :
Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because :
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Boron is unable to form $\mathrm{BF}_{6}^{3-}$.
Reason (R) : Size of B is very small.
In the light of the above statements, choose the correct answer from the options given below :
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Activated charcoal adsorbs SO2 more efficiently than CH4.
Reason R : Gases with lower critical temperatures are readily adsorbed by activated charcoal.
In the light of the above statements, choose the correct answer from the options given below.
Given below are two statements.
Statement I: The chlorides of $\mathrm{Be}$ and $\mathrm{Al}$ have Cl-bridged structure. Both are soluble in organic solvents and act as Lewis bases.
Statement II: Hydroxides of $\mathrm{Be}$ and $\mathrm{Al}$ dissolve in excess alkali to give beryllate and aluminate ions.
In the light of the above statements, choose the correct answer from the options given below.
Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula?
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion : Boric acid is a weak acid.
Reason R : Boric acid is not able to release $\mathrm{H}^{+}$ ion on its own. It receives $\mathrm{OH}^{-}$ ion from water and releases $\mathrm{H}^{+}$ ion.
In the light of the above statements, choose the most appropriate answer from the options given below.
Match List - I with List - II.
| List I (Processes / Reactions) |
List II (Catalyst) |
||
|---|---|---|---|
| (A) | 2SO$_2$(g) + O$_2$(g) $ \to $ 2SO$_3$(g) | (I) | Fe(s) |
| (B) | 4NH$_3$(g) + 5O$_2$(g) $ \to $ 4NO(g) + 6H$_2$O(g) | (II) | Pt(s) $ - $ Rh(s) |
| (C) | N$_2$(g) + 3H$_2$(g) $ \to $ 2NH$_3$(g) | (III) | V$_2$O$_5$ |
| (D) | Vegetable oil(l) + H$_2$ $ \to $ Vegetable ghee(s) | (IV) | Ni(s) |
Choose the correct answer from the options given below :
Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "$X$" with 6-equivalents of "$Y$". "$X$" and "$Y$", respectively are :
The total number of acidic oxides from the following list is
NO, N2O, B2O3, N2O5, CO, SO3, P4O10