Match the species in Column I with the bond order in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | NO | 1. | 1.5 |
| B. | CO | 2. | 2.0 |
| C. | O$$_2^-$$ | 3. | 2.5 |
| D. | O$$_2$$ | 4. | 3.0 |
A. $\rightarrow$ (3)
B. $\rightarrow$ (4)
C. $\rightarrow(1)$
D. $\rightarrow(2)$
$$ \begin{gathered} \text { A. } N O(7+8=15)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi \star 2 p_x^1 \\ \text { Bond order }=\frac{1}{2}\left(N_b-N_a\right)=\frac{10-5}{2}=2.5 \end{gathered}$$
B. $\mathrm{CO}(6+8=14)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$
Bond order $=\frac{10-4}{2}=3$
C. $\mathrm{O}_2^{-}(8+8+1=17)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma * 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^2 \approx \pi * 2 p_y^1$
Bond order $=\frac{10-7}{2}=1.5$
D. $\mathrm{O}_2(8+8=16)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^{\star} 2 p_x^1 \approx \pi^{\star} 2 p_y^1$
Bond order $=\frac{10-6}{2}=2$
Match the items given in Column I with examples given in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Hydrogen bond | 1. | C |
| B. | Resonance | 2. | LiF |
| C. | Ionic solid | 3. | H$$_2$$ |
| D. | Covalent solid | 4. | HF |
| 5. | O$$_3$$ |
A. $\rightarrow$ (4)
B. $\rightarrow$ (5)
C. $\rightarrow$ (2)
D. $\rightarrow$ (1)
A. Hydrogen bond $\rightarrow \mathrm{HF}$
B. Resonance $\rightarrow \mathrm{O}_3$
C. Ionic bond $\rightarrow$ LiF
D. Covalent solid $\rightarrow \mathrm{C}$
Match the shape of molecules in Column I with the type of hybridisation in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Tetrahedral | 1. | sp$$^2$$ |
| B. | Trigonal | 2. | sp |
| D. | Linear | 4. | sp$$^3$$ |
A. $\rightarrow(3)$
B. $\rightarrow(1)$
C. $\rightarrow(2)$
A. Tetrahedral shape $-s p^3$ hybridisation
B. Trigonal shape $-s p^2$ hybridisation
C. Linear shape - sp hybridisation
Assertion (A) Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R) This is because sodium and chloride ions acquire octet in sodium chloride formation.
Assertion (A) Though the central atom of both $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ molecules are $s p^3$ hybridised, yet $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle is greater than that of $\mathrm{H}-\mathrm{O}-\mathrm{H}$.
Reason (R) This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.