| For the following molecules: \[PC{{l}_{5}},Br{{F}_{3}},ICl_{2}^{-},XeF_{5}^{-},NO_{3}^{-},Xe{{O}_{2}}{{F}_{2}},PCl_{4}^{+},CH_{3}^{+}\] |
| Calculate the value of \[\frac{(a+b)}{c}\] |
| a = Number of species having \[s{{p}^{3}}d\]-hybridisation |
| b = Number of species which are planar |
| c = Number of species which are non-planar |
A) 2
B) 4
C) 3
D) 5
Correct Answer: C
Solution :
| \[PC{{l}_{5}}\xrightarrow{{}}s{{p}^{3}}d,\]non-planer |
| \[Br{{F}_{3}}\xrightarrow{{}}s{{p}^{3}}d\], bent, T-shape, planar |
| \[ICl_{2}^{-}\xrightarrow{{}}s{{p}^{3}}d\], linear, planar |
| \[XeF_{5}^{-}\xrightarrow{{}}s{{p}^{3}}{{d}^{3}}\], pentagonal planar |
| \[NO_{3}^{-}\xrightarrow{{}}s{{p}^{2}}\], planar |
| \[Xe{{O}_{2}}{{F}_{2}}\xrightarrow{{}}s{{p}^{3}}d\], see-saw, non-planar |
| \[PCl_{4}^{+}\xrightarrow{{}}s{{p}^{3}}\], tetrahedral, non-planar |
| \[CH_{3}^{+}\xrightarrow{{}}s{{p}^{2}}\]. Trigonal planar |
| \[a=4,\,\,b=5,\,\,c=3\] |
| So, \[\frac{a+b}{c}=3\] |
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