InterviewSolution
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InterviewSolution
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Assign the position of the element having outer electronic configuration a.`ns^(2)np^(4)` for `n=3` b. `(n-1)d^(2)ns^2` for `n=4` and c. `(n-2) f^(7)(n-1)d^(1)ns^(2)` for `n=6`, in the periodic table. |
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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 + number of d–block groups + number of p–electrons = 2 + 10 + 4 = 16 Therefore, the element belongs to the `3^(rd)` period and `16^(th)` group of the periodic table. Hence, the element is Sulphur. (ii) Since n = 4, the element belongs to the `4^(th)` 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 + number of d–block groups = 2 + 2 = 4 Therefore, it is a `4^(th)` period and `4^(th)` group element. Hence, the element is Titanium. (iii) Since n = 6, the element is present in the `6^(th)` period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] `4f^(7) 5d^(1) 6s^(2)`. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium. |
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