The oxidation numbers of nitrogen in nh3, hno3, and no2 are, respectively: a) -3, -5, +4 b) +3, +5, +4 c) -3, +5, -4 d) -3, +5, +4 is it d?

Taking into account the definition of oxidation number, the correct answer is the option D: oxidation numbers of nitrogen in NH₃, HNO₃, and NO₂ are -3, +5, +4.

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

To determine the oxidation state of different elements it is necessary to know that:

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

In this case, the oxidation numbers of nitrogen in each case is calculated as:

oxidation numbers of nitrogen + 3= 0

oxidation numbers of nitrogen = -3

oxidation numbers of nitrogen + 1 - 6= 0

oxidation numbers of nitrogen - 5 = 0

oxidation numbers of nitrogen = +5

oxidation numbers of nitrogen - 4= 0

oxidation numbers of nitrogen = +4

Finally, the correct answer is the option D: oxidation numbers of nitrogen in NH₃, HNO₃, and NO₂ are -3, +5, +4.

Learn more about the oxidation number:

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Explanation:

1) NH₃

Since N is more electronegative than H, the oxidation state of H is positive, So, it is +1.

Using the rule that the sum of the oxidation states of all the atoms is equal to the charge of the compound, and since NH₃ is a neutral compound (net charge equals 0), you can set the following equation for the oxidation states:

2) HNO₃

In acids the oxidation state of hydrogen is +1 and the oxidation state of oxygen is -2.

So, you can set this valid equation for the neutral compound:

3) NO₂

The oxidation state for oxygen in any compound (except in peroxides) is -2, so:

Summarizing, the oxidation numbers of nitrogen in the three given compounds are:  - 3, +5, and + 4. That is the option D).

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.