Consider the following compounds:
(i)`IF_(5)` (ii)`ClI_(4)^(-)` (iii)`XeO_(2)F_(2)` (iv)`NH_(2)^(-)` (v) `BCl_(3)` (vi)`BeCl_(2)` (vii)`AsCl_(4)^(+)` (viii)`B(OH)_(3)` (ix)`NO_(2)^(-)` (x)`ClO_(2)^(+)`
Then, calculate value of "`x+y-z`", here x, y and z are total number of compounds in given compounds in which central atom used their all three p-orbitals, only two p-orbitals and only one p-orbital in hybridisation respectively:

Answer

8

Explanation

`(i) IF_(5)(sp^(3)d) " " (ii) ClI_(4)^(-)(sp^(3)d^(2)) " " (iii) XeO_(2)F_(2)(sp^(3)d) " " (iv) NH_(2)^(-)(sp^(3))`
`(v) BCl_(3)(sp^(2)) " " (vi) BeCl_(2)(sp) " " (vii) AsCl_(4)^(+)(sp^(3)) " " (viii)B(OH)_(3)(sp^(2))`
(ix) `NO_(2)^(-)(sp^(2)) " " (x) ClO_(2)^(+)(sp^(2))`
`x=(sp^(3)2+sp^(3)d(1)+sp^(3)d^(2)(2)=5`
y=4, z=1, 5+4-1=8