Top 25 Repeated JEE Mains Questions to Score 90+ in Chemistry

Top 5 High weightage Chapters in JEE Main Chemistry

Some chapters contribute to a large portion of the questions asked every year in JEE Main. Students can study these high weightage chapters to score maximum marks by studying only 5-6 chapters from the entire JEE Main Chemistry syllabus. Given below 6 chapters based on previous year question analysis, that are most important and frequently asked in JEE Main.

Most Asked Topics in JEE Main Chemistry from Last 5 Years

To score maximum marks it is important to focus more on the specific topics that are asked in JEE Main.These topics are those from which JEE Main Chemistry repeated questions are asked, instead of studying the entire chapter prioritizing the topics mentioned below in the form of table:

Top 25 Most Repeated JEE Main Chemistry Questions

The Top 25 repeated JEE Mains Questions to Score 90+ in Chemistry are based on concepts that appear repeatedly in previous year exams. These questions are highly recommended and mostly based on NCERT that helps students to score good marks in JEE main. Refer to the Top 25 repeated JEE Mains questions in Chemistry given below:

Question 1: Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula $\mathrm{CrCl}_3 .3 \mathrm{NH}_3 .3 \mathrm{H}_2 \mathrm{O}$ reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is__________

(1) 0

(2) 1

(3) 2

(4) 3

Answer:

Given,

The complex $\mathrm{CrCl}_3 .3 \mathrm{NH}_3 .3 \mathrm{H}_2 \mathrm{O}$ produces 3 moles AgCl when 1 mole of the complex is treated with excess $\mathrm{AgNO}_3$.

This means that all 3 chloride ions are present outside the co-ordination sphere and satisfy only the primary valency. The given complex is, therefore,

$\mathrm{CrCl}_3 .3 \mathrm{NH}_3 .3 \mathrm{H}_2 \mathrm{O}$

Thus, no chloride ion satisfies the secondary valency (co-ordination number)

Hence, the answer is option (1).

Question 2: Complex X of composition $\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6 \mathrm{Cl}_n$ has a spin-only magnetic moment of 3.83 BM. It reacts with $\mathrm{AgNO}_3$ and shows geometrical isomerism. The IUPAC nomenclature of X is:

(1) Tetraaquadichlorido chromium (IV) chloride dihydrate

(2) Hexaaqua chromium (III) chloride

(3) Dichloridotetraaqua chromium (IV) chloride dihydrate

(4) Tetraaquadichlorido chromium (III) chloride dihydrate

Answer:

$\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6 \mathrm{Cl}_n ; \quad\left(\mu_{\text {complex }}\right)$ spin $=3.8$ B.M

From data of magnetic moment oxidation number of Cr should be +3
Hence complex is $\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6 \mathrm{Cl}_3$̣

Complex shows geometrical isomerism therefore formula of the complex is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} .2 \mathrm{H}_2 \mathrm{O}$

Its IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate

Hence, the answer is the option(4).

Question 3: The correct order of energy of absorption for the following metal complexes is

$\mathrm{A}:\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+}, \mathrm{B}:\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}, \mathrm{C}:\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

(1) $\mathrm{C}<\mathrm{B}<\mathrm{A}$

(2) $\mathrm{B}<\mathrm{C}<\mathrm{A}$

(3) $\mathrm{C}<\mathrm{A}<\mathrm{B}$

(4) $\mathrm{A}<\mathrm{C}<\mathrm{B}$

Answer:

Stronger the ligands the splitting will be larger. If splitting is large the energy of absorption will be high.

Increasing field strength : $\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{en}$

So, the energy of absorption : $\mathrm{C}<\mathrm{B}<\mathrm{A}$

Hence, the answer is the option (1).

Question 4: For a $\mathrm{d}^4$ metal ion in an octahedral field, the correct electronic configuration is:

(1) $\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^1$ when $\Delta_0<\mathrm{P}$

(2) $\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^1$ when $\Delta_0>\mathrm{P}$

(3) $\mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^0$ when $\Delta_0<\mathrm{P}$

(4) $\mathrm{t}_{2 \mathrm{~g}}^2 \mathrm{e}_{\mathrm{g}}^2$ when $\Delta_0<\mathrm{P}$

Answer:

We know

P = Pairing energy

$\Delta_0$ = Crystal field splitting energy

(i) If $\Delta_0<\mathrm{I}$, the fourth electron enters one of the $\mathrm{e}_{\mathrm{g}}$ orbitals giving the configuration $t_{2 g}^3 e_g^1$

(ii) If $\Delta_0>\mathrm{P}$ it becomes more energetically favourable for the fourth electron to occupy an orbital with configuration $t_{2 g}^4 e_g^0$

Hence, the answer is the option (1).

Question 5: The group having triangular planar structures is :

(1) $B F_3, N F_3, C O_3^{2-}$

(2) $\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}, \mathrm{SO}_3$

(3) $\mathrm{NH}_3, \mathrm{SO}_3, \mathrm{CO}_3^{2-}$

(4) $\mathrm{NCl}_3, \mathrm{BCl}_3, \mathrm{SO}_3$

Answer:

The group having triangular planar structures is $\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}, \mathrm{SO}_3$

In $\mathrm{CO}_3^{2-}$ ion, the C atom is $s p^2$ hybridized. This results in a triangular planar structure.

In $\mathrm{NO}_3^{-}$ ion, the N atom is $s p^2$ hybridized. This results in a triangular planar structure.

In $\mathrm{SO}_3$ S atom is $s p^2$ hybridized. This results in a triangular planar structure.

In all the above molecules/ions, the central atom has 3 bonding domains and a bond angle of 120o .

Hence, the answer is an option (2).

Question 6: $s p^3 d^2$ hybridization is not displayed by :

(1) $\mathrm{BrF}_5$

(2) $\mathrm{SF}_6$

(3) $\left[\mathrm{CrF}_6\right]^{3-}$

(4) $\mathrm{PF}_5$

Answer:

$\mathrm{sp}^3 \mathrm{~d}^2$ hybridization is not displayed by $\mathrm{PF}_5$. The central P atom has 5 sets of electron pairs (all of which are bonding) hence it shows $\mathrm{sp}^3 \mathrm{~d}$ hybridization. $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization corresponds to 6 sets of electron pairs.

Rest all other options display $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization.

Hence, the answer is option (4).

Question 7: Which one of these factors does not effect the stability of conformation in acyclic compounds?

(1) Steric interaction

(2) Torsional strain

(3) Electrostatic forces of interaction

(4) Angle strain

Answer:

Angle strain:- Angle strain is the increase in the potential energy of a molecule due to bond angles deviating from the ideal values. eg: Cyclopropane.

Angle strain is not found in acyclic compounds, it is found in cyclic compounds.

Hence, the answer is the option (4).

Question 8: Among the following molecules/ions, $C_2^{2-}, N_2^{2-}, O_2^{2-}, O_2$ which one is diamagnetic and has the shortest bond length?

(1) $\mathrm{O}_2$

(2) $N_2^{2-}$

(3) $O_2^{2-}$

(4) $C_2^{2-}$

Answer:

$C_2^{2-}$:- No of electrons = 14

$\sigma_{1 S}^2 \sigma_{1 S^2}^* \sigma_{2 S}^2 \sigma_{2 S^2}^* \pi_{2 p x}^2=\pi_{2 p y}^2 \sigma_{2 p z^2}$

(1) It is diamagnetic

(2)Bond order $=\frac{10-4}{2}=3$

$O_2^{2-}$:- No of electrons=18

$\sigma_{1 S}^2 \sigma_{1 S^2}^* \sigma_{2 S}^2 \sigma_{2 S^2}^* \sigma 2 p z^2 \pi 2 p x^2=\pi_{2 p y}^2 \pi_{2 p x^2}^*=\pi_{2 p y^2}^*$

(1) It is diamagnetic

(2)Bond Order $=\frac{10-8}{2}=1$

$\because$ B.O $\propto \frac{1}{\text { Bond length }}$

∴ The correct answer is $\mathrm{C}_2^{2-}$

Hence, the answer is the option (4).

Question 9: The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 faraday .The amount of $\mathrm{PbSO}_4$ electrolyzed in g during the process is _________

(Molar mass of $\mathrm{PbSO}_4=303 \mathrm{~g} \mathrm{~mol}^{-1}$)

(1) 22.8

(2) 15.2

(3) 7.6

(4) 11.4

Answer:

As we have learned,

When the anodic half of the lead storage battery is recharged, the following reaction occurs

$\mathrm{PbSO}_4+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}+\mathrm{SO}_4^{2-}$

Given that the total charge passed is 0.05 F

Now, from the balanced equation, it is clear that

2 F charge electrolysis 1 mole of $\mathrm{PbSO}_4$

$\therefore 0.05 \mathrm{~F}$ charge electrolysis 0.025 mole of $\mathrm{PbSO}_4$

Thus, the mass of $\mathrm{PbSO}_4$ electrolysed $=0.025 \times 303=7.575 \mathrm{~g}$

Hence, the answer is the option (3).

Question 10: Match List - I with List - II :

Choose the most appropriate answer from the options given below :

(1) $(a)-(i),(b)-(i v),(c)-(i i i),(d)-(i i)$

(2) $(a)-(i i i),(b)-(i),(c)-(i i),(d)-(i v)$

(3) $(a)-(i i i),(b)-(i),(c)-(i v),(d)-(i i)$

(4) $(a)-(i i),(b)-(i),(c)-(i i i),(d)-(i v)$

Answer:

The correct combination of the parameters and their units are given below:

(a) Cell constant $\left(\mathrm{G}^*\right): \mathrm{m}^{-1}(\mathrm{iii})$

(b) Molar conductivity $\left(\wedge_{\mathrm{m}}\right): \mathrm{Scm}^2 \mathrm{~mol}^{-1}(\mathrm{i})$

(c) Conductivity $(\kappa): \Omega^{-1} \mathrm{~m}^{-1}(\mathrm{iv})$

(d) Degree of dissociation $(\alpha):$ dimensionless(ii)

Hence, the answer is the option (3).

Question 11: Calculate the standard cell potential (in V) of the cell in which following reaction takes place:

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given that

$\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0=\mathrm{xV}$

$\mathrm{E}_{\mathrm{Fe}^{2+}+\mid \mathrm{Fe}}^0=\mathrm{yV}$

$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}}^0=\mathrm{z} \mathrm{V}$

(1) $\mathrm{x}-\mathrm{z}$

(2) $x-y$

(3) $x+2 y-3 z$

(4) $x+y-z$

Answer:

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given,

$\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0=\mathrm{xV}$

$\mathrm{E}_{\mathrm{Fe}^{2+}+\mid \mathrm{Fe}}^0=\mathrm{yV}$

$\mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}}^0=\mathrm{zV}$

The given equation can be represented in the form of a cell as

$\mathrm{Fe}^{2+}\left|\mathrm{Fe}^{+3} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$

Standard EMF of given cell reaction

$$
\mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0-\mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}^{2+}}^0
$$

It is evident that in order to find the above cell potential, we need to find the electrode potential of the $\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}$ couple.

In order to calculate the value of $\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+},}^0$,, we need to use the given values of $\mathrm{E}_{\mathrm{Fe}^{2+} \mid \mathrm{Fe}^2}^0$ and $\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{F}}^0$.

Now,

$$
\mathrm{Fe}^{+2}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \quad \mathrm{E}^0=\mathrm{y}, \Delta \mathrm{G}^0=-2 \mathrm{Fy}
$$

$$
\mathrm{Fe}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}^0=\mathrm{z}, \Delta \mathrm{G}^0=-3 \mathrm{Fz}
$$

$\mathrm{Fe}^{+3}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2} ; \Delta \mathrm{G}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

Thus we can write

$-1 \times \mathrm{F} \times \mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}^{2+}}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=3 \mathrm{z}-2 \mathrm{y}$

$\mathrm{E}_{\text {cell }}^0=\mathrm{x}-(3 \mathrm{z}-2 \mathrm{y})=\mathrm{x}+2 \mathrm{y}-3 \mathrm{z}$

Hence, the answer is the option (3).

Question 12: The correct order of the oxidation states of nitrogen in $\mathrm{NO}, \mathrm{N}_2 \mathrm{O}, \mathrm{NO}_2$ and $\mathrm{N}_2 \mathrm{O}_3$ is :

(1) $\mathrm{NO}_2<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{N}_2 \mathrm{O}$

(2) $\mathrm{NO}_2<\mathrm{N}_2 \mathrm{O}_3<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}$

(3) $\mathrm{N}_2 \mathrm{O}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{NO}<\mathrm{NO}_2$

(4) $\mathrm{N}_2 \mathrm{O}<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{NO}_2$

Answer:

The oxidation state of Nitrogen in

$\begin{aligned} & \mathrm{NO} \rightarrow+2 \Rightarrow \mathrm{~N}+(-2)=0 \Rightarrow \mathrm{~N}=+2 \\ & \mathrm{~N}_2 \mathrm{O} \rightarrow+1 \Rightarrow 2 \mathrm{~N}-2=0 \Rightarrow \mathrm{~N}=+1 \\ & \mathrm{NO}_2 \rightarrow+4 \Rightarrow \mathrm{~N}-2 \times 2=0 \Rightarrow \mathrm{~N}=+4 \\ & \mathrm{~N}_2 \mathrm{O}_3 \rightarrow+3 \Rightarrow 2 \mathrm{~N}-2 \times 3=0 \Rightarrow \mathrm{~N}=+3\end{aligned}$

Order would be

$\mathrm{N}_2 \mathrm{O}<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{NO}_2$

Hence, the answer is the option (3).

Question 13: $\Delta_{\mathrm{f}} \mathrm{G}^0$ at 500 K for substance ‘S’ in liquid state and gaseous state are +100.7 kcal mol −1 and +103 kcal mol−1 , respectively. Vapour pressure (in atm) of liquid ‘S’ at 500 K is approximately equal to : (R=2 cal K−1 mol−1 ):

(1) 0.1

(2) 1

(3) 10

(4) 100

Answer:

Δ G of equilibrium

$\Delta \mathrm{G}_0=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{c}}$

At Equilibrium

$\Delta G=0$

and $Q=K_c$

$\Delta \mathrm{G}_{\text {Reaction }}^{\mathrm{o}}=103-100.7$

$=2.3 \mathrm{kcal}=2.3 \times 10^3 \mathrm{cal}$

$\Delta G^o=-2.303 R T \log K_p 2.3 \times 10^3$

$=-2.303 \times 2 \times 500 \log \mathrm{~K}_{\mathrm{p}} \log \mathrm{K}_{\mathrm{p}}=-1 \mathrm{~K}_{\mathrm{p}}=10^{-1}=0.1 \mathrm{~atm}$

Hence, the answer is an option (1).

Question 14: The combustion of benzene (l) gives $\mathrm{CO}_2$(g) and $\mathrm{H}_2 \mathrm{O}$(l). Given that heat of combustion of benzene at constant volume is $-3263.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $258^{\circ} \mathrm{C}$ heat of combustion (in kJ mol−1) of benzene at constant pressure will be :

$\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.$)

(1) -3267.6

(2) 4152.6

(3) -452.46

(4) 3260

Answer:

We know this formula -

$d H=d E+n R d T$

$\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)$

$\Delta n_g=6-\frac{15}{2}$

For chemical reactions;

$\Delta H=\Delta U+\Delta n_g R T \quad \therefore \Delta U=Q_V$ and $\Delta H=Q_P$

$\Delta H=-3263.9 \times 1000+\left(-\frac{3}{2}\right) \times 8.314 \times 298$

$\Delta H=(-3263.9 \times 1000-3716.36) J \Delta H=-3267.6 K J$

Hence, the correct answer is option (1).

Question 15: At 320 K, a gas $A_2$ is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in $\mathrm{Jmol}^{-1}$ is approximately : (R=8.314 $J K^{-1} \mathrm{~mol}^{-1}$ ; ln 2=0.693; ln 3=1.098)

(1) 4763

(2) 2068

(3) 1844

(4) 4281

Answer:

As we learnt in

Relation between Gibbs energy and reaction Quotient -

$\triangle G=\triangle G^{\circ}+R T l_n Q$

$\triangle G^{\circ}$ is standard Gibbs energy.

$Q \rightarrow$ Reaction Quotient

$A_2(g) \rightleftharpoons 2 A(g) 1-0.2 \quad 2 \times 0.20 .8 \quad 0.4$

$K_p=\frac{\left(P_A^2\right)}{\left(P_{A_2}\right)}=\frac{0.4 \times 0.4}{0.8}=0.2$

$\Delta G^{\circ}=-2.303 \times 8.314 \times 320 \log 0.2=4281 \mathrm{~J} / \mathrm{mole}$

Hence, the answer is an option (4).

Question 16: The absolute value of the electron gain enthalpy of halogens satisfies :

(1) Cl> F > Br > I

(2) F >Cl> Br > I

(3) Cl> Br > F > I

(4) I > Br >Cl> F

Answer:

The absolute value of the electron gain enthalpy of halogens satisfies:

Cl > F > Br > I

Chlorine has a higher electron gain enthalpy than fluorine due to less electron density.

Hence, the answer is the option (1).

Question 17: HBr reacts fastest with :

(1) Propan-2-01

(2) 2-methyl-propan-1-01

(3) 2- methyl-propan-2-01

(4) Propan-1-01

Answer:

As we learn

Physical properties of Phenol -

– Phenols are more acidic than phenol.

– They are less soluble in water but readily soluble in organic solvents.

– They liquify due to high hygroscopic nature.

2-methyl-2-propanol will react fastest with HBr.

Hence, the answer is the option (3).

Question 18: Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Cis form of alkene is found to be more polar than the trans form

Reason (R): Dipole moment of trans isomer of 2-butene is zero.

In the light of the above statements, choose the correct answer from the options given below :

(1) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

(2) (A) is true but (R) is false

(3) Both (A) and (R) are true and (R) is the correct explanation of (A)

(4) (A) is false but (R) is true

Answer:

Dipole moment is a vector quantity and for compound net dipole moment is the vector sum of all dipoles hence dipole moment of cis form is greater than trans form.

Hence, the answer is the option (3).

Question 19: Given below are two statements :

Statement I: Hyperconjugation is a permanent effect.

Statement II : Hyperconjugation in ethyl cation $\left(\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}_2\right)$ involves the overlapping of $\mathrm{C}_{\mathrm{sp}}{ }^2-\mathrm{H}_{1 \mathrm{~s}}$ bond with empty 2 p orbital of other carbon.

Choose the correct option :

(1) Both statement I and statement II are false

(2) Statement I is false but statement II is true

(3) Statement I is true but statement II is false

(4) Both Statement I and statement II are true.

Answer:

Hyperconjugation is a permanent effect.

In the ethyl cation, hyperconjugation involves overlapping of $\mathrm{sp}^3$ orbital with vacant p-orbital of the carbocation.

Thus, statement I is true while statement II is false.

Hence, the answer is the option (3).

Question 20: Match List I with List II

Choose the correct answer from the options given below:

(1) A-II, B-III, C-I, D-IV

(2) A-III, B-I, C-IV, D-II

(3) A-III, B-IV, C-I, D-II

(4) A-IV, B-III, C-I, D-II

Answer:

(A) $\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{NH}_2 \& \mathrm{C}_{\mathrm{N}} \mathrm{H}-\mathrm{C}-\mathrm{C}$ : functional isomer(III)

(B) Metamer

(C) Tautamer (IV)

(D) Position isomer (II)

Hence, the answer is the option (2)

Question 21: The number of P−O bonds in $\mathrm{P}_4 \mathrm{O}_6$ is:

(1) 6

(2) 9

(3) 12

(4) 18

Answer:

As we learned in

Phosphorus pentaoxide:

$\mathrm{P}_4 \mathrm{O}_{10}$, a white solid, formed when $P_4$ is burned in excess of air

- wherein

$P_4+5 O_2 \rightarrow P_4 O_{10}$

Phosphorus trioxide

$\mathrm{P}_4 \mathrm{O}_6$, dimer, white solid, obtained by burning phosphorus in a limited amount of air

- wherein

$\mathrm{P}_4+3 \mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_6$

The number of P-O bonds in $\mathrm{P}_4 \mathrm{O}_6$ is 12.

Hence, the answer is an option (3).

Question 22: When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ’X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is :

(1) Fe

(2) Zn

(3) Ca

(4) Al

Answer:

As we have learnt,

Aluminium dissolves in NaOH to form a gelatinous white precipitate of $\mathrm{Al}(\mathrm{OH})_3$ and liberates $\mathrm{H}_2$ gas

$\mathrm{Al}(\mathrm{OH})_3$ that dissolves more than NaOH forming Sodium Aluminate.

$\mathrm{Al}(\mathrm{OH})_3+\mathrm{NaOH} \rightarrow \underset{\text { soluble }}{\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]}$

$\mathrm{Al}(\mathrm{OH})_3$ gives Alumina on heating which is used as an adsorbent in Chromatography.

$\mathrm{Al}(\mathrm{OH})_3 \xrightarrow{\Delta} \mathrm{Al}_2 \mathrm{O}_3$

Hence, the answer is an option (4).

Question 23: Among the following allotropic forms of sulphur, the number of allotropic forms, which will show para magnetism is ______

(A) $\alpha$ - sulphur (B) $\beta$ - sulphur (C) $\mathrm{S}_2$ - form

Answer:

$\mathrm{S}_2$ is like $\mathrm{O}_2$ i;e paramagnetic as per molecular orbital theory.

$\alpha-$ sulphur and $\beta-$ sulphur both are diamagnetic.

So, only $\mathrm{S}_2$-form is paramagnetic.

Hence, the answer is (1).

Question 24: Match List - I with List - II

Choose the correct answer from the options given below :

(1) (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)

(2) (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)

(3) (a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)

(4) (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)

Answer:

1. Haber's process is used for $\mathrm{NH}_3$ synthesis.

$\mathrm{N}_2+3 \mathrm{H}_2 \stackrel{\mathrm{Fe} / \mathrm{Mo}}{\rightleftharpoons} 2 \mathrm{NH}_3$

2. Ostwald's process is used for $\mathrm{HNO}_3$ synthesis using a Pt catalyst.

3. Contact process is used for $\mathrm{H}_2 \mathrm{SO}_4$ synthesis using $\mathrm{N}_2 \mathrm{O}_5$ catalyst.

4. In the Hall-Heroult process, electrolytic reduction of impure alumina can be done. (Aluminium extraction).

Hence, the answer is the option (1).

Question 25: In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively :

(1) 33 and 25

(2) 33 and 75

(3) 50 and 75

(4) 67 and 75

Answer:

As we have learned,

Graphite has $\mathrm{sp}^2$ hybridization

$\% \mathrm{p}-$ character $=\frac{2}{3} \times 100=67 \%$

Diamond has $\mathrm{sp}^3$ hybridisation

$\% \mathrm{p}-$ character $=\frac{3}{4} \times 100=75 \%$

Hence, the answer is the option (4).