The electronic configuration of an element is `1s^(2)2s^(2)2p^(6),3s^(2)3p^(5)`. The atomic number of element present just below the above element in periodic table is:
A. 33
B. 34
C. 36
D. 49
Correct Answer - A
The element which present just below the given element will have outermost electromic configuration as `4s^(2) 4p^(3),` so its full electronic configuration is `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),4s^(2), 3d^(10), 4p^(3)` and hence, its atoimc number is 33.
Correct Answer - D
As ionisation energy decreases electropositive character increases and thus the tendency to form cation increases. Therefore, metallic character increases.
Correct Answer - B
As ionisation energies decrease down the group the reducing character increases and as across the period , ionisation energies increases, so reducing character, decreases.
(a) The electronic configuration of element `X` is `[Rn]^(86) 5f^(14) 6d^(10) 7s^(2)`
(b) If belongs to d-block as last electron enters in d subshell.
(c ) As number of electrons...
Correct Answer - C
Silver belongs to `V^(th)` period. So the atomic number of element placed above and below will be `47 - 18 = 29` and `47 +32 = 79`...
Correct Answer - C
The metallic character of the elements is highest at the extremely left (low ionisation energies) and then decreases across the period from left to right (ionisation energies...