Question

Sulphur, nitrogen and fluorine form a compound (X) that contain 31.1% sulphur and 55.33% fluorine (by mass) respectively. If molar mass of compound is 103 g/mol then select the correct statement(s) about (X).

Total number of lone pair in 1 molecule of 'X' is 10

Hybridisation of central atom in 'X' is sp3

X is planar molecule

1 molecule of 'X' contains four sigma bond​

Answer 1

Answer:

1, 2 and 4th option are correct

Explanation:

please find the attached file

Answer 2

Answer:

The given compound is SF₃N. The statements 1st, 2nd and 4th are correct about this compound.

Explanation:

From the given data, we can find the empirical formula of the compound (X)

S = 31.1%, F= 55.33%,  ∴ N= 100-(31.3+55.33)= 13.57%

 $Element\\ \\ Sulphur\\ \\ Fluorine\\ \\ Nitrogen$             $Percentage\\ \\ 31.1\\ \\ 55.33\\ \\ 13.57$           $Atomic \\ mass\\ \\ 32\\ \\ 19\\ \\ 14$            $Relative \\ moles\\ \\ 0.971\\ \\ 2.912\\ \\ 0.969$           $Simple\\ ratio\\ \\ 1\\ \\ 3\\ \\1$

• Therefore, the empirical formula of compound is SF₃N .
• The formula mass of the compound= 32+3(19)+14 = 103g

   ∴               The molecular formula is also SF₃N.

• From the structure (attached figure)

       Hybridization = lone pair on central atom + bond pair

                               =0 + 4 =4

Therefore, hybridization of compound is sp³.

• The structure is tetrahedral, not planar.
• Total four sigma bonds formed by sulphur, three with fluorine and one with nitrogen.
• Total lone pairs in the molecule= 3(no. of F)+ no. of nitrogen

                                               lone pairs =3(3)+1 =10

Therefore, 1st, 2nd and 4th statement are correct.