Name the element having general electronic
configuration ns²np⁴ in fourth period.
Answers
Answer:
ANSWER
(i) Since n = 3, the element belongs to the 3
rd
period. It is a p–block element since the last electron occupies the p–orbital.
There are four electrons in the p–orbital. Thus, the corresponding group of the element
= Number of s–block groups(3s
2
) + number of d–block groups([Ne]
10
+ number of p–electrons(3p
4
)
= 2 + 10 + 4
= 16
Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur([Ne]
10
3s
2
3p
4
)
(ii) Since n = 4, the element belongs to the 4th period. It is a d–block element as d–orbitals are incompletely filled.
There are 2 electrons in the d–orbital.
Thus, the corresponding group of the element
= Number of s–block groups(4s
2
) + number of d–block groups(3d
2
)
= 2 + 2
= 4
Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium([Ar]
18
3d
2
4s
2
.
(3)(n−1)d
1
ns
2
For n = 6
= (6−1)d
1
6s
2
= 5d
1
6s
2
Now if we write it in the order of filling the sub shells, we know that first 6s fill then 6d fill so
6s
2
5d
1
Now
period = Highest principal quantum number = 6th period
Group = 1 + 2 = 3 group
And the atomic number is 57
Answer:
Hyyy
Explanation:
(i) Since n = 3, the element belongs to the 3rd
period. It is a p–block element since the last electron occupies the p–orbital.
There are four electrons in the p–orbital. Thus, the corresponding group of the element= Number of s–block groups(3s² ) + number of d–block groups([Ne] ^10 + number of p–electrons(3p ^4)
= 2 + 10 + 4
= 16
Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur([Ne] ^10 3s² 3p^4)