Explanation
= $$0.125$$
So, the correct option is $$B$$
The key feature Of Bohr's theory Of spectrum Of hydrogen atom is the quantization Of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy Of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.A diatomic molecule has moment Of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is
As we know,
$$r \propto n^2$$
$$E \propto -\dfrac{z^2}{n^2}$$
$$V\propto \dfrac{z}{n}$$
$$mur=nh^2 \pi$$
So, independent of $$z$$ so it will remain same.
So, orbital angular momentum of electron
Pottassium selenate is isomorphous to $$K_2SO_4$$ and thus its molecular formula is $$K_2SeO_4$$
Now mol.wt of $$K_2SeO_4=(39\times 2+a+4\times 16)$$
$$\Rightarrow 142+a$$
Where $$'a'$$ is at.wt of $$Se$$
$$(142+a)g\;K_2SeO_4$$ $$=Se\ a\ g$$
100g $$K_2SeO_4=\dfrac{a\times 100}{142+a}$$
$$\%$$ of $$Se=50$$
$$a=142$$
$$\approx 142$$
Also equivalent weight of $$K_2SeO_4=\dfrac{Mol.wt}{2}$$
$$\Rightarrow \dfrac{2\times 39+142+64}{2}$$
$$\Rightarrow 142$$
Hence $$(A)$$ is the correct answer.
$$MC{{L}_{2}}$$ have $$ 50$$ atomic number so $$ M$$ will be $$ 16$$ atomic number.
$$ C{{l}_{2}}=17+17 $$
atomic number is $$34+16=50$$.
And molecule will be $$SC{{l}_{2}}$$ and according to this shape of the molecule is bent.
And $$S$$ have two lone pair.
The elements in which the last electron enters the outermost s-orbital are called s-block elements.
The s-block elements have two groups ($$1$$ and $$2$$).
Thus, s-block elements include lithium $$(Li)$$ and magnesium $$(Mg).$$
The periodic table shows exactly where these elements are within the s-block.
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