The valency of nitrogen in nitrogen dioxide is:
A. 1
B. 2
C. 3
D. 4
Answer
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Hint: Nitrogen dioxide is a gas with no net charge on the molecule. We know that valency of oxygen is 2. Write down the formula of nitrogen dioxide and calculate the valency of nitrogen to balance the valency over two atoms of oxygen in nitrogen dioxide.
Complete step by step solution:
Nitrogen dioxide is written as: \[N{O_2}\]. Oxygen has a valency of 2 and if we see on the periodic table, it is also electronegative out of the two. Valency of the element tells about the combining power of the element. Since the oxygen needs two more electrons to gain stability, its oxidation state will be -2. Since the net charge on the molecule is zero, we will assume the valency of nitrogen is $x$. Writing the balance equation to bring the net oxidation state of nitrogen dioxide as zero, we get,
$x + 2 \times ( - 2) = 0 \Rightarrow x = 4$. Thus the oxidation state of nitrogen is +4 and valency of nitrogen is 4. We can say that option D is correct.
Note: It is always the best way to first find the most electronegative of the elements comprising the compound and then equate the sum of the individual oxidation states to the net oxidation state of the molecule. This builds with practice but little memory in the beginning.
Complete step by step solution:
Nitrogen dioxide is written as: \[N{O_2}\]. Oxygen has a valency of 2 and if we see on the periodic table, it is also electronegative out of the two. Valency of the element tells about the combining power of the element. Since the oxygen needs two more electrons to gain stability, its oxidation state will be -2. Since the net charge on the molecule is zero, we will assume the valency of nitrogen is $x$. Writing the balance equation to bring the net oxidation state of nitrogen dioxide as zero, we get,
$x + 2 \times ( - 2) = 0 \Rightarrow x = 4$. Thus the oxidation state of nitrogen is +4 and valency of nitrogen is 4. We can say that option D is correct.
Note: It is always the best way to first find the most electronegative of the elements comprising the compound and then equate the sum of the individual oxidation states to the net oxidation state of the molecule. This builds with practice but little memory in the beginning.
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