What is the maximum number of electrons that can be associated with the following quantum numbers n 3 L 1m =- 1?

The electrons identified by quantum numbers n and l:(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3 , l = 1

Can be placed in order of increasing energy as

• (c) < (d) < (b) < (a) 

• (d) < (b) < (c) < (a)

• (b) < (d) < (a) < (c)

• (b) < (d) < (a) < (c)

B.

(d) < (b) < (c) < (a)

a.) (n + l) = 4 + 1 = 5 b.) (n + l) = 4 + 0 = 4 c.) (n + 1) = 3 + 2 = 5 d.) (n + 1) = 3 + 1 = 4Higher the value of (n+1), higher the energy. if (n+1) are same, sub orbit with a lower value of n has lower energy. Thus,

3p<4s<3d<4p

a.) (n + l) = 4 + 1 = 5 b.) (n + l) = 4 + 0 = 4 c.) (n + 1) = 3 + 2 = 5 d.) (n + 1) = 3 + 1 = 4Higher the value of (n+1), higher the energy. if (n+1) are same, sub orbit with a lower value of n has lower energy. Thus,

3p<4s<3d<4p

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What is the maximum number of electrons that can be associated with the following set of quantum numbers?n = 3, l=1 and m= 1

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Text Solution

42106

Answer : B

Solution : n=3, l=1, the resultant subshell is 3p. Thus, number of orbitals having n=3, l=1 and m=-1 is one and the maximum number of electrons that can be accommodated in this orbital is two.