Match List I with List II .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top LIST I (Molecule) LIST II (Shape) A. NH_3 I. Square pyramid B. BrF_5 II. Tetrahedral C. PCl_5 III. Trigonal pyramidal D. CH_4 IV. Trigonal bipyramidal Choose the correct answer from the options given below:

Chemical Bonding and Molecular Structure PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q8 with Solution

Match List I with List II

Choose the correct answer from the options given below:

View written solutionFree

Correct answer: B

A. $\mathrm{NH_3}$
- Central atom: N
- Bond pairs = 3, lone pairs = 1
- Electron-pair geometry = tetrahedral
- Molecular shape = trigonal pyramidal
- So, A $\to$ III
B. $\mathrm{BrF_5}$
- Central atom: Br
- Bond pairs = 5, lone pairs = 1
- Total electron pairs = 6
- Electron-pair geometry = octahedral
- Molecular shape = square pyramidal
- So, B $\to$ I
C. $\mathrm{PCl_5}$
- Central atom: P
- Bond pairs = 5, lone pairs = 0
- Molecular shape = trigonal bipyramidal
- So, C $\to$ IV
D. $\mathrm{CH_4}$
- Central atom: C
- Bond pairs = 4, lone pairs = 0
- Molecular shape = tetrahedral
- So, D $\to$ II

$A\to III, \quad B\to I, \quad C\to IV, \quad D\to II$

This corresponds to Option B.

Stored correct answer = B

My derived answer = B

So, they agree.