- 1(current)
Q. No.Match the element in Column I with that in Column II.
| Column-I | Column-II | ||
| (a) | Copper | (i) | Non-metal |
| (b) | Fluorine | (ii) | Transition metal |
| (c) | Silicon | (iii) | Lanthanoid |
| (d) | Cerium | (iv) | Metalloid |
| (a) | (b) | (c) | (d) | |
| 1. | (ii) | (iv) | (i) | (iii) |
| 2. | (ii) | (i) | (iv) | (iii) |
| 3. | (iv) | (iii) | (i) | (ii) |
| 4. | (i) | (ii) | (iii) | (iv) |
The period and group number of the element with Z =114 are:
| 1. | 8th period and 16th group | 2. | 7th period and 14th group |
| 3. | 14th period and 7th group | 4. | 7th group and 14th period |
Sodium generally does not show an oxidation state of +2, because of:
1. High first ionization potential.
2. High second ionization potential.
3. Large ionic radius.
4. High electronegativity.
The electronic configuration that represents the d-block element is:
1. 1s2 2s2 2p6 3s23p63d10 4s24p6
2. 1s2 2s2 2p6 3s23p63d10 4s24p1
3. 1s2 2s2 2p6 3s23p63d10 4s2
4. 1s2 2s2 3s23p6 4s2
The general outer electronic configuration of s, p, d , and f-block elements respectively would be :
| 1. | ns1-2, nd2np1-6, (n-1)d1-10np0-2, (n-2)f1-14(n-1)d0-10ns2 |
| 2. | ns1-2, ns2np1-6, (n-1)f1-10ns0-2, (n-2)g1-14(n-1)d0-1ns2 |
| 3. | ns1-2, ns2np1-6, (n-1)d1-10ns1-2, (n-2)f1-14(n-1)d0-1 ns2 |
| 4. | np1-2, nd2np1-6, (n-1)d1-10ns0-2, (n-2)f1-14(n-1)d0-10ns2 |
Consider the following electronic configuration of an element (P) :
The correct statement about element 'P' is :
1. It belongs to the 6th period and the 1st group.
2. It belongs to the 6th period and the 2nd group.
3. It belongs to the 6th period and the 3rd group.
4. None of the above.
- 1(current)
Q. No.
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