JEE Advance - Chemistry (2012 - Paper 1 Offline - No. 3)
The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table?
$${}_{29}^{63}Cu$$ + $${}_1^1H$$ $$\to$$ $$6{}_0^1n$$ + $${}_2^4\alpha $$ + 2$${}_1^1H$$ + X
$${}_{29}^{63}Cu$$ + $${}_1^1H$$ $$\to$$ $$6{}_0^1n$$ + $${}_2^4\alpha $$ + 2$${}_1^1H$$ + X
Answer
8
Explanation
Considering the reaction (replace $\alpha$-particle by ${ }_2^4 \mathrm{He}$ )
$$ { }_{29}^{63} \mathrm{Cu}+{ }_1^1 \mathrm{H} \rightarrow 6{ }_0^1 n+{ }_2^4 \mathrm{He}+2{ }_1^1 \mathrm{H}+\mathrm{X} $$
Equating mass numbers on both the sides, we get
$$ 63+1=1 \times 6+4 \times 1+1 \times 2 + X$$
$$ \Rightarrow $$ $$X=64-12=52 $$
Equating atomic numbers on both the sides, we get
$$ 29+1=6 \times 0+2+2 \times 1+\mathrm{Y} $$
$$\Rightarrow \mathrm{Y}=30-4=26 $$
So, the element is ${ }_{26}^{52} \mathrm{Fe}$ and iron is a $d$-block element, which belongs to Group 8 of the periodic table.
$$ { }_{29}^{63} \mathrm{Cu}+{ }_1^1 \mathrm{H} \rightarrow 6{ }_0^1 n+{ }_2^4 \mathrm{He}+2{ }_1^1 \mathrm{H}+\mathrm{X} $$
Equating mass numbers on both the sides, we get
$$ 63+1=1 \times 6+4 \times 1+1 \times 2 + X$$
$$ \Rightarrow $$ $$X=64-12=52 $$
Equating atomic numbers on both the sides, we get
$$ 29+1=6 \times 0+2+2 \times 1+\mathrm{Y} $$
$$\Rightarrow \mathrm{Y}=30-4=26 $$
So, the element is ${ }_{26}^{52} \mathrm{Fe}$ and iron is a $d$-block element, which belongs to Group 8 of the periodic table.
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