A group 15 element forms $\mathrm{d} \pi-\mathrm{d} \pi$ bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form $d \pi-d \pi$ bond.
The atomic number of the element is _______ .
Explanation:
Phosphorus, an element from group 15 of the periodic table, can form a $d \pi-d \pi$ bond with transition metals. Among the hydrides of group 15, phosphine ($\text{PH}_3$) is considered the strongest base, except for ammonia ($\text{NH}_3$).
Number of oxygen atoms present in chemical formula of fuming sulphuric acid is ___________.
Explanation:
Fuming sulfuric acid, often known as oleum, can be represented chemically using the formula $ H_2S_2O_7 $. This compound is formed by adding excess sulfur trioxide ($ SO_3 $) to sulfuric acid ($ H_2SO_4 $). The chemical reaction can be represented as follows:
$ H_2SO_4 + SO_3 \rightarrow H_2S_2O_7 $
From the formula $ H_2S_2O_7 $, to determine the number of oxygen atoms, we look directly at the subscript corresponding to oxygen in the chemical formula. In this case, the subscript is 7, which indicates that there are 7 oxygen atoms present in the molecule of fuming sulfuric acid.
$\mathrm{Cl}_2 \mathrm{O}_7, \mathrm{CO}, \mathrm{PbO}_2, \mathrm{~N}_2 \mathrm{O}, \mathrm{NO}, \mathrm{Al}_2 \mathrm{O}_3, \mathrm{SiO}_2, \mathrm{~N}_2 \mathrm{O}_5, \mathrm{SnO}_2$
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.
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 Oâ‚…)
y = 4 (from (OH)â‚„)
z = 8 (from 8Hâ‚‚O)
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
Consider the following sulphur based oxoacids.
$\mathrm{H}_{2} \mathrm{SO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}$ and $\mathrm{H}_{2}\mathrm{S}_{2} \mathrm{O}_{7}$.
Amongst these oxoacids, the number of those with peroxo $(\mathrm{O}-\mathrm{O})$ bond is ________.
Explanation:
The number of non-ionisable protons present in the product B obtained from the following reactions is ___________.
$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{PCl}_{3} \rightarrow \mathrm{C}_{2} {\mathrm{H}_{5} \mathrm{Cl}+\mathrm{A}} $
$\mathrm{A}\,{+}\, \mathrm{PCl}_{3} \rightarrow \mathrm{B}$
Explanation:
$ \underset{\text{(A)}}{\mathrm{H}_{3} \mathrm{PO}_{3}}+\mathrm{PCl}_{3} \rightarrow \underset{\text{(B)}}{\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}} $
Structure of $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}$
Total 2 non-ionizable protons are present
Consider the following reactions:
PCl3 + H2O $\to$ A + HCl
A + H2O $\to$ B + HCl
The number of ionisable protons present in the product B is _______________.
Explanation:
$ \underset{\mathrm{A}}{\mathrm{P}(\mathrm{OH})_{2} \mathrm{Cl}_{2}}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\text{B}}{\mathrm{P}(\mathrm{OH})_{2} \mathrm{Cl}}+\mathrm{HCl} $
Hydrogen attached with oxygen are ionisable. Hence number of ionisable protons present in compound $B$ are 2.
Explanation:
HgS, Sb2S3 are insoluble in 50% HNO3
So, answer is 4.
Explanation:
F cannot go in +5 oxidation state.
At is radioactive.
The number of halogen forming halic (V) acid
HClO3
HBrO3
HIO3
So answer is 3
Explanation:
XeF6 + H2O $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Hydrolysis}^{Partial}} $ XeO2F2 + 2HF
In XeF6, central atom Xe has one lone pair all 6 fluorine have 3 lone pairs each.
So, total number of lone pair on XeF6 = 1 + (6 $\times$ 3) = 19
Explanation:
(A) $\alpha$-sulphur
(B) $\beta$-sulphur
(C) S2-form
Explanation:
The values of 'a' is _______. (Integer answer)
Explanation:
${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$ (Reduction)
${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$
_____________________________________
$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$
For balancing in basic medium, Add an equal number of OH$-$ that of H+, we get
$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$
$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$
or ${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$ .... (i)
On comparing (i) with
${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$
We get, a = 12; b = 4; c = 2; d = 6
Dissolving $1.24 \mathrm{~g}$ of white phosphorous in boiling $\mathrm{NaOH}$ solution in an inert atmosphere gives a gas $\mathbf{Q}$. The amount of $\mathrm{CuSO}_{4}$ (in g) required to completely consume the gas $\mathbf{Q}$ is _________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{P}=31, \mathrm{~S}=32, \mathrm{Cu}=63$ ]
Explanation:
$\mathop {{P_4}}\limits_{\matrix{ {1.24\,g} \cr {or} \cr {0.01\,mole} \cr } } +3NaOH+3H_2O\to PH_3+3NaH_2PO_2$
As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P4 is limiting)
$\mathop {2P{H_3}}\limits_{0.01\,mole} +3CuSO_4\to Cu_3P_2+3H_2SO_4$
Amount of CuSO4 required = $\frac{3\times0.01}{2}$ mole
Mass of CuSO4 (in g) required = $\frac{0.03}{2}\times(63+32+16\times4)$
$=\frac{0.03}{2}\times159=2.38$ g
Explanation:
B2H6 and B3N3H6 have polar bond, but do not have same kind of atom.
Explanation:
$Xe{F_4} + {O_2}{F_2}\buildrel {143\,K} \over \longrightarrow \mathop {Xe{F_6}}\limits_{(Y)} + {O_2}$
The structure of XeF6 is
Y compound (XeF6) has 3 lone pair in each fluorine and one lone pair in xenon.
Hence, total number of lone pairs electrons is 19.
${N_2}{O_3},{N_2}{O_5},$ ${P_4}{O_6},{P_4}{O_7},$ ${H_4}{P_2}{O_5},{H_5}{P_3}{O_{10}},$ ${H_2}{S_2}{O_3},{H_2}{S_2}{O_5}$
Explanation:
The structures of various molecules given in problem are discussed below -
1. N2O3 It is the tautomeric mixture of following two structures -
Conclusion 1 bridging oxo group is present in the compound.
2. N2O5 It has following structure.
Conclusion 1 bridging oxo group is present in the compound.
where, pi = initial pressure, pt = total pressure, y = number of gaseous products per mole of reactant
3. P4O6
Conclusion 6 bridging oxo groups are present in the compound.
4. P4O7
Conclusion 6 bridging oxo groups are present in the compound.
5. H4P2O5
Conclusion 1 bridging oxo group is present in the compound.
6. H5P3O10
Conclusion 2 bridging oxo groups are present in the compound.
7. H2S2O3
Conclusion This compound does not contain any bridging oxo group.
8. H2S2O5
Conclusion This compound also does not contain any bridging oxo group.
Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is ____________.
Explanation:
The reaction is
B2H6 + 6CH3OH $\to$ 2B(OCH3)3 + 6H2
From the reaction, 1 mol of B2H6 reacts with 6 mol of CH3OH to produce 2 mol of B(OCH3)3.
Therefore, 3 mol of B2H6 would react with 18 mol of CH3OH to produce 6 mol of B(OCH3)3.
The number of lone pairs of electrons in N2O3 is ___________.
Explanation:
The structure of N2O3 is
Therefore, the total number of lone pairs is 8.
Among the following, the number of compounds that can react with PCl5 to given POCl3 is _____________.
O2, CO2, SO2, H2O, H2SO4, P4O10
Explanation:
The following reactions show how PCl5 reacts with different compounds to form POCl3 :
$CO_2 + PCl_5 \rightarrow POCl_3 + COCl_2$
$H_2O + PCl_5 \rightarrow POCl_3 + 2HCl$
$SO_2 + PCl_5 \rightarrow POCl_3 + SOCl_2$
$P_4O_{10} + 6PCl_5 \rightarrow 10POCl_3$
$H_2SO_4 + PCl_5 \rightarrow POCl_3 + HSO_3Cl + HCl$
From these reactions, it is clear that CO2, H2O, SO2, P4O10, and H2SO4 can react with PCl5 to give POCl3.
There are 5 compounds that meet this criterion.
The value of $n$ in the molecular formula $\mathrm{Be_n Al_2Si_6O_{18}}$ is ___________.
Explanation:
In the given molecular formula $\mathrm{Be_n Al_2Si_6O_{18}}$, according to charge balance in a molecule, we get
$2n+2(+3)+6(+4)-18(2)=0$
$\Rightarrow n=3$
So, the formula is $\mathrm{Be_3 Al_2Si_6O_{18}}$.
The total number of diprotic acids among the following is:
H3PO4, H2SO4, H3PO3, H2CO3, H2S2O7, H3BO3, H3PO2, H2CrO4 and H2SO3
Explanation:
A diprotic acid is an acid that contains within its molecular structure two hydrogen atoms per molecule capable of dissociating (i.e., ionisable protons) in water.
$ \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7, \mathrm{H}_2 \mathrm{CrO}_4, \mathrm{H}_3 \mathrm{PO}_3, \mathrm{H}_2 \mathrm{SO}_3 $
Structure of each of the compounds is given below :
(A) Calculate the amount of calcium oxide required to react with 852 g of P$_4$O$_{10}$.
(B) Write the structure of P$_4$O$_{10}$.
Explanation:
(A) The balanced chemical reaction between CaO and P$_4$O$_{10}$ is shown below.
$\mathrm{6CaO+P_4O_{10}\rightarrow 2Ca_3(PO_4)_2}$
Given, mass of $\mathrm{P_4O_{10}=852~g}$
The first step is to calculate the molar mass of $\mathrm{P_4O_{10}}$. The molar mass can be used to calculate the number of moles present in the given sample of $\mathrm{P_4O_{10}}$.
Molar mass of
$\mathrm{P_4O_{10}}=4\times31+10\times16=284$ g
Moles of $\mathrm{P_4O_{10}}=\frac{\mathrm{Mass~of~P_4O_{10}}}{\mathrm{Molar~mass~of~P_4O_{10}}}=\frac{852}{284}$
Moles of $\mathrm{P_4O_{10}}$ = 3 moles
1 mol of $\mathrm{P_4O_{10}}$ reacts with 6 mol of CaO to produce 2 mol of Ca$_3$(PO$_4$)$_2$.
$\therefore$ 3 mol of $\mathrm{P_4O_{10}}$ will react with 3 $\times$ 6 = 18 mol of CaO.
Mass of 1 mol of CaO = 40 + 16 = 56 g
$\therefore$ Mass of 18 mol of CaO = 18 $\times$ 56 = 1008 g
(B) Structure of $\mathrm{P_4O_{10}}$
In $\mathrm{P_4O_{10}}$, eqach P atom is bonded to 4 O atoms such that there are three P-O single bonds and one P-O double bond. Two phosphorous atoms share a single O atom.
Final Answer
(A) 1008 g
(B)