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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
The number of species having non-pyramidal shape among the following is ___________.

(A) SO3

(B) NO$$_3^ -$$

(C) PCl3

(D) CO$$_3^{2 - }$$

## Explanation

Hence, non-pyramidal species are SO3, NO$$_3^{- }$$ and CO$$_3^{2 - }$$.
2

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
AB3 is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is __________. (Integer answer)

## Explanation

T-shaped molecule means 3 sigma bond and 2 lone pairs of electron on central atom.

3

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
The total number of electrons in all bonding molecular orbitals of $$O_2^{2 - }$$ is ______________.

(Round off to the nearest integer)

## Explanation

Nb = No of electrons in bonding molecular orbital

Na $$=$$ No of electrons in anti bonding molecular orbital

(1) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10

(2) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -

Here Na = 10

and Nb = 10

In O atom 8 electrons present, so in O2, 8 $$\times$$ 2 = 16 electrons present.

Then in $$O_2^ +$$ no of electrons = 15

in $$O_2^ -$$ no of electrons = 17

in $$O_2^{2 - }$$ no of electrons = 18

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ *$$

$$\therefore\,\,\,\,$$ Nb = 10

and Na = 8
4

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
The difference between bond orders of CO and NO$$^ \oplus$$ is $${x \over 2}$$ where x = _____________. (Round off to the Nearest Integer)

## Explanation

Bond order of CO = 3

Bond order of NO+ = 3

Difference = 0 = $${x \over 2}$$

$$\Rightarrow$$ x = 0

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