The correct order of melting points of hydrides of group 16 elements is:

H_{2} S > H_{2} S e > H_{2} T e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} S < H_{2} S e < H_{2} T e

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} S > H_{2} S e < H_{2} T e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} S < H_{2} S e > H_{2} T e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $H_{2} S < H_{2} S e < H_{2} T e$
For covalent hydrides,
$M e l t i n g p o i n t \propto V a n d e r w a a l^{'} s f o r c e \propto M o l e c u l a r m a s s$
Thus, melting point order is $H_{2} T e > H_{2} S e > H_{2} S$ .

Suggest Corrections

Similar questions

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

H_{2} O < H_{2} T e < H_{2} S e < H_{2} S

H_{2} O > H_{2} T e > H_{2} S e > H_{2} S

H_{2} O > H_{2} S > H_{2} S e > H_{2} T e

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

H_{2} E

(E = O, S, S e, T e a n d P o

I

H_{2} S, H_{2} S e

H_{2} T e

I I

H_{2} S, H_{2} S e

H_{2} T e

Join BYJU'S Learning Program