Among the following if the number of linear molecules are x and the number of angular molecules are y. Then the value of x/y would be
$O_{3} . N_{2} O, N O_{2}, S O_{2}, C O_{2}, S n C l_{2}, C l_{2} O, H C N, + N O_{2}$

Open in App

Solution

Liner olecule have $s p$ hybridisation
$N_{2} O, C O_{2}, H C N$ and $+ N O_{2}$ are $s p$ hybridised. Hence, linear in shape.
Hence, $x = 4$
$O_{3} = s p^{2} h y b r i d i s e d$ and angular in shape.
$N O_{2} = s p^{2} h y b r i d i s e d$ and angular in shape.
$S O_{2} = s p^{2} h y b r i d i s e d$ and angular in shape.
$S n C l_{2} = s p^{3} h y b r i d i s e d$ with one lone pair and hence, angular in shape.
$C l_{2} O$ = $s p^{3} h y b r i d i s e d$ with one lone pair and hence, angular in shape. are angular.
Hence $y = 5$
Hence, x = 4 and y = 5 so $\frac{x}{y} = \frac{4}{5} = 0.8$

Suggest Corrections

Similar questions

x

y

x \times y

O_{3}, N_{2} O, N O_{2}, S O_{2}, C O_{2}, S n C l_{2}, C l_{2} O, H C N, + N O_{2}

O_{3}, N_{2} O, N O_{2}, S O_{2}, C O_{2}, S n C l_{2}, C l_{2} O, H C N, + N O_{2}

x

y

\frac{x}{y}

S O_{2}, C O_{2}, N_{2} O, O F_{2}, {I C l_{2}}^{-}, {I_{3}}^{-}, N O_{2}, {N O_{2}}^{-}, {N O_{2}}^{+}, {B F_{2}}^{-}, B e C l_{2}, {N_{3}}^{-}, H C N .

P_{4} O_{10}, C O, S O_{2}, C l O_{2}, C l_{2} O, N_{2} O, N O_{2}, C l_{2} O_{6}, C O_{2}, C l_{2} O_{7}, S O_{3}

{B F}_{3}, {C O}_{2}, {S O}_{2}, {P C I}_{5}, {C I F}_{3}, {N H}_{3}, {C H}_{4}

Join BYJU'S Learning Program