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