Question

Match the atomic number of the elements given in column I with the periods given in column II and mark the appropriate choice.

COLUMN I(Atomic Number)	COLUMN II(Period)
(A)- 31	(i)- 5
(B)- 50	(ii)- 3
(C)- 56	(iii)- 4
(D)- 14	(iv)- 6

(A)- $(A)\to (i)\,,\,(B)\to (ii)\,,\,(C)\to (iii)\,,\,(D)\to (iv)$
(B)- $(A)\to (ii),\,(B)\to \left( i \right),\,(C)\to (iv),\,(D)\to (iii)$
(C)- $(A)\to (iii),\,(B)\to (iv),\,(C)\to (i),\,(D)\to (ii)$
(D)- $(A)\to (iii),\,(B)\to (i),\,\,(C)\to (iv),\,(D)\to (ii)$

Answer 1

Hint : To answer this question the range of atomic numbers of each period should be remembered. Each period starts with alkali metals and ends with noble gas.
The modern periodic table consists of 7 periods and 18 groups. The properties of elements along a period shows a gradual change whereas the properties of elements along the group are the same.

Complete step by step solution :
Let’s look the answer to this question:
Elements in a periodic table are arranged in increasing order of atomic number. The modern periodic table is based on modern periodic law which states that, “The properties of elements are a periodic function of their atomic numbers.”
This means that after certain intervals the properties of the element repeat due similar outer electronic configuration.

The range of atomic numbers in period 3 is from 11 to 18. So, element having atomic number 14 belongs to period 3.
Hence, $(D)\to \left( ii \right)$
The range of atomic numbers in period 4 is from 19 to 36. So, element having atomic number 31 belongs to period 4.
Hence, $(A)\to \left( iii \right)$
The range of atomic numbers in period 5 is from 37 to 54. So, element having atomic number 50 belongs to period 5.
Hence, $(B)\to \left( i \right)$
The range of atomic numbers in period 6 is from 55 to 86. So, element having atomic number 56 belongs to period 6.
Hence, $\left( C \right)\to \left( iv \right)$
Now, option (D) contains all the relations correctly.

So, the correct answer is “Option D”.

Note : While answering this type of question students should eliminate the wrong options in order to reach the correct option faster. Elements of the same period differ in properties while elements of the same group have similar properties.