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Question

Write the electronic configurations of the following ions:
(a) H
(b) Na+
(c) O2
(d) F

(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s1
(b) 2p3 and
(c) 3p5 ?

Which atoms are indicated by the following configurations?
(a) [He] 2s1
(b) [Ne] 3s23p3
(c) [Ar] 4s23d1

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Solution

(i).

(a) H
The electronic configuration of H atom is 1s1.
A negative charge on the species indicates the gain of an electron by it. Electronic configuration of H=1s2

(b) Na+
The electronic configuration of Na atom is 1s22s22p63s1.
A positive charge on the species indicates the loss of an electron by it. therefore Electronic configuration of Na+=1s22s22p63s0 or 1s22s22p6

(c) O2
The electronic configuration of 0 atom is 1s22s22p4.
A dinegative charge on the species indicates that two electrons are gained by it.
Electronic configuration of O2 ion = 1s22s2p6

(d) F
The electronic configuration of F atom is 1s22s22p5.
A negative charge on the species indicates the gain of an electron by it.
Electronic configuration of F ion = 1s22s22p6

(ii).

(a) 3s1
Completing the electron configuration of the element as
1s22s22p63s1.
Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
Atomic number of the element = 11

(b) 2p3
Completing the electron configuration of the element as
1s22s22p3.
Number of electrons present in the atom of the element
= 2 + 2 + 3 = 7
Atomic number of the element = 7

(c) 3p5
Completing the electron configuration of the element as
1s22s22p5.
Number of electrons present in the atom of the element
= 2 + 2 + 5 = 9
Atomic number of the element = 9

(iii).

(a) [He] 2s1
The electronic configuration of the element is [He] 2s1=1s22s1.
Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).

(b) [Ne] 3s23p3
The electronic configuration of the element is [Ne] 3s23p3=1s22s22p63s23p3.
Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s2 3p3 is phosphorus (P).

(c) [Ar] 4s23d1
The electronic configuration of the element is [Ar] 4s23d1=1s22s22p63s23p64s23d1.
Atomic number of the element = 21 Hence, the element with the electronic configuration [Ar] 4s23d1 is scandium (Sc).


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