JEE Chemistry 2024 Question with Solution

Given: The IUPAC name of the element is Unununnium.

Find: The value of $n$, the group number of the element.

The systematic name is formed from the digits of the atomic number.

• un $\rightarrow$ $1$
• un $\rightarrow$ $1$
• un $\rightarrow$ $1$

So the atomic number is

$$Z = 111$$

Thus the element is roentgenium.

To determine the group, use its electronic configuration. A possible configuration is

$$[Rn] \, 7s^2 \, 5f^{14} \, 6d^9$$

For a $d$-block element, the group number is the sum of electrons in $ns$ and $(n-1)d$.

$$\text{Group Number} = (\text{electrons in } ns) + (\text{electrons in } (n-1)d)$$

Substituting the values,

$$\text{Group Number} = 2 + 9 = 11$$

Alternatively, with the more stable configuration

$$[Rn] \, 7s^1 \, 5f^{14} \, 6d^{10}$$

we get

$$\text{Group Number} = 1 + 10 = 11$$

Therefore, the element belongs to Group $11$.

So, the value of $n$ is $11$.

Given: The element name is Unununnium.

Find: The group number.

The numerical roots used in IUPAC temporary naming are:

• $0 = \text{nil}$
• $1 = \text{un}$
• $2 = \text{bi}$
• $3 = \text{tri}$
• $4 = \text{quad}$
• $5 = \text{pent}$
• $6 = \text{hex}$
• $7 = \text{sept}$
• $8 = \text{oct}$
• $9 = \text{enn}$

In Unununnium, the roots are un-un-un, so the atomic number is $111$.

Element with $Z = 111$ lies in the $d$-block of the $7$th period. Its group is obtained from the outer configuration by adding the $s$ and $d$ electrons involved in bonding position.

Using

$$[Rn] \, 7s^2 \, 5f^{14} \, 6d^9$$

we get

$$11 = 2 + 9$$

Hence the element is in Group $11$.

Therefore, the required value is $11$.