
Equilateral shape has
(A) sp hybridisation
(B) $ s{{p}^{2}} $ hybridisation
(C) $ s{{p}^{3}} $ hybridisation
(D) $ ds{{p}^{2}} $ hybridisation
Answer
507.6k+ views
Hint :To solve this question, we need to look for the angles associated with each type of hybridisation. We need to check each option separately and evaluate the angles of the hybridisations.
Complete Step By Step Answer:
In the question, we need to find the angle associated with each of the above hybridisations and then eliminate the one that does not match our requirements. According to the question, we have an equilateral shape, that is the angle must be of $ 120{}^\circ $ . Options not following this requirement will be eliminated.
Let us start with the first option, $ sp $ hybridisation has a shape of linear. And in the case of linear, the bond angle is $ 180{}^\circ $ and that is not equal to the condition in the question which is required. This is why we have to move on to the next option eliminating the first option.
In the second option, we have $ s{{p}^{2}} $ hybridisation. And such $ s{{p}^{2}} $ hybridisation have the shape of trigonal planar and the bond angle between such is $ 120{}^\circ $ so this is the correct option. Moving on to the next option, the bond angle between molecules of $ s{{p}^{3}} $ hybridisation is $ 109{}^\circ $ . And the molecules following the hybridisation of the last option have a bond angle of $ 90{}^\circ $ and $ 120{}^\circ $ .
So the correct option is option number (B).
Note :
The $ s{{p}^{3}} $ hybridisation has a tetrahedral shape and so it has a unique angle. And the molecules having hybridisation of $ ds{{p}^{2}} $ has trigonal bi-pyramidal shape and so it has two angles, in for the planar bonds and other for the perpendicular bonds.
Complete Step By Step Answer:
In the question, we need to find the angle associated with each of the above hybridisations and then eliminate the one that does not match our requirements. According to the question, we have an equilateral shape, that is the angle must be of $ 120{}^\circ $ . Options not following this requirement will be eliminated.
Let us start with the first option, $ sp $ hybridisation has a shape of linear. And in the case of linear, the bond angle is $ 180{}^\circ $ and that is not equal to the condition in the question which is required. This is why we have to move on to the next option eliminating the first option.
In the second option, we have $ s{{p}^{2}} $ hybridisation. And such $ s{{p}^{2}} $ hybridisation have the shape of trigonal planar and the bond angle between such is $ 120{}^\circ $ so this is the correct option. Moving on to the next option, the bond angle between molecules of $ s{{p}^{3}} $ hybridisation is $ 109{}^\circ $ . And the molecules following the hybridisation of the last option have a bond angle of $ 90{}^\circ $ and $ 120{}^\circ $ .
So the correct option is option number (B).
Note :
The $ s{{p}^{3}} $ hybridisation has a tetrahedral shape and so it has a unique angle. And the molecules having hybridisation of $ ds{{p}^{2}} $ has trigonal bi-pyramidal shape and so it has two angles, in for the planar bonds and other for the perpendicular bonds.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

