Among the following species, identify the isostructural pairs
${NF}_{3}, {NO}_{3}^{-}, {BF}_{3}, H_{3} O^{+}, {HN}_{3}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$[N F_{3}, N O_{3}^{-}] a n d [B F_{3}, H_{3} O^{+}]$

$[N F_{3}, H N_{3}] a n d [N O_{3}^{-}, B F_{3}]$

$[N F_{3}, H_{3} O^{+}] a n d [N O_{3}^{-}, B F_{3}]$

$[N F_{3}, H_{3} O^{+}] a n d [H N_{3}, B F_{3}]$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

For species to be isostructural they should have same hybridized state same number of b.p. and l.p.

$\begin{array}{l} {NF}_{3}; H = \frac{1}{2} (5 + 3) = 4 \Rightarrow N is {sp}^{3} hybridized; 1 lone-pair \\ {NO}_{3}; H = \frac{1}{2} (5 + 1) = 3 \Rightarrow N is {sp}^{2} hybridized; 0 lone - pairs \\ {BF}_{3} H = \frac{1}{2} (3 + 3) = 3 \Rightarrow B is {sp}^{2} hybridized; 0 lone - pairs \\ H_{3} O^{+} H = \frac{1}{2} (6 + 3 - 1) = 4 \Rightarrow O is {sp}^{3} hybridized; 1 lone - pair \end{array}$

These three ${sp}^{2} - hybrid$ orbitals are attached to each other trigonally with an angle of $120^{0}$ and they are overlapped with three p-orbitals of three F-atoms on their axes. Hence the geometry of $B F_{3}$ molecule is trigonal planar.

Thus isostructural pairs are $[N F_{3}, H_{3} O^{+}] a n d [N O_{3}^{-}, B F_{3}]$