Question

41. Elements X and Y of 3rd period form binary
compounds with fluorine P and Q respectively. Some
information is given below.
Compound P : Non-polar and has only 6 bond
angles of 90° each.
Compound Q : Non-polar and has only 12 bond
angles of 90° each.
The number of lone pairs present in each molecule
of compounds P and Q are 'a' and 'b' respectively.
The value of (a + b) is
45

Answer 1

Answer:

33

Explanation:

A molecule whose shape is Trigonal bipyramdial has 6 90° bond angle

since third period is mentioned , lets assume the molecule $PF_5$(non polar) .

When we calculate the no. of non bonding lone pair electrons(which only F atoms possess in $PF_5$) comes out to be

3*5=15=a

A molecule whose shape is octahedral has 12 90° bond angle

since third period is mentioned , lets assume the molecule $SF_6$(non-polar)

When we calculate the no. of non bonding lone pair electrons(which only F atoms possess in $SF_6$) comes out to be

3*6=18=b

So, a+b

    =15+18

    =33

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