Question

Match List-I with List-II
List-I List-II
(Species) (Hybrid orbitals)
(a) $S{F}_4$ (i) $s{p}^3{d}^2$

(b) $I{F}_5$ (ii) $d^{2}s{p}^3$

(c) $N{O}_2^{+}$ (iii) $s{p}^{3}d$

(d) $N{H}_4^{+}$ (iv) $s{p}^3$

(v) $sp$

Choose the correct answer from the options given below :

• A. $\left(a\right)-\left(ii\right),\left(b\right)-\left(i\right),\left(c\right)-\left(iv\right),\left(d\right)-\left(v\right)$
• B. $\left(a\right)-\left(iv\right),\left(b\right)-\left(iii\right),\left(c\right)-\left(ii\right),\left(d\right)-\left(v\right)$
• C. $\left(a\right)-\left(i\right),\left(b\right)-\left(ii\right),\left(c\right)-\left(v\right),\left(d\right)-\left(iii\right)$
• D. $\left(a\right)-\left(iii\right),\left(b\right)-\left(i\right),\left(c\right)-\left(v\right),\left(d\right)-\left(iv\right)$

Answer 1

The correct option is D $\left(a\right)-\left(iii\right),\left(b\right)-\left(i\right),\left(c\right)-\left(v\right),\left(d\right)-\left(iv\right)$

$S{F}_4$- S is surrounded by 4 sigma bond pair and 1 lone pair. Thus, its steric number is 5. So it is $s{p}^{3}d\text{hybridised.}$

$I{F}_5$- I is surrounded by 5 sigma bond pair and 1 lone pair. Thus, its steric number is 6. So, it is $s{p}^3{d}^2\text{hybridised.}$

$N{O}_2^{+}$- N has 2 sigma bond pair and no lone pair. Thus, its steric number is 2. So, it is $sp\text{hybridised.}$

$N{H}_4^{+}$- N has 4 sigma bond pair and no lone pair. Thus, its steric number is 4. So it is $s{p}^3\text{hybridised.}$

Hence, option (d) is correct.