Ncert Exercises & Solutions

The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+l), lower is the energy. For orbitals having same values of (n+l), the orbital with lower value of n will have lower energy.

I. Based upon the above information, arrange the following orbitals in the increasing order of energy.

1. 1s, 2s, 3s, 2p

2. 4s, 3s, 3p, 4d

3. 5p, 4d, 5d, 4f, 6s

4. 4f, 6d, 7s, 7p

II. Based upon the above information. Solve the questions given below.

1. Which of the following orbitals has the lowest energy?

4d, 4f, 5s, 5p

2. Which of the following orbitals has the highest energy?

5p, 5d, 5f, 6s, 6p

I. (a) (n+l) values of 1s=1+0=1, 2s=2+0=2, 3s=3+0=3, 2p=2+1=3

Hence, increasing order or their energy is

1s<2s<2p<3s

(b) 4s=4+0=4, 3s=3+0=3, 3p=3+1=4, 4d=4+2=6

Hence, 3s<3p<4s<4d

(c) 5p=5+1=6, 4d=4+2=6, 5d=5+2=7, 4f=4+3=7, 6s=6+0=6

Hence, 4d<5p<6s<4f<5d

(d) 5f=5+3=8, 6d=6+2=8, 7s=7+0=7, 7p=7+1=8.

Hence, 7s<5f<6d<7p

II. (a) (n+l) values of 4d=4+2=6, 4f=4+3=7, 5s=5+0=5, 7p=7+1=8

Hence, 5s has the lowest energy.

(b) 5p=5+1=6, 5d=5+2=7, 5f=5+3=8, 6s=6+0=6, 6p=6+1=7

Hence, 5f has highest energy.

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