Answer:
bent
Explanation:
The chlorite ion has been pictured in the image attached with its two main resonance contributors.
The ion has four regions of electron density as shown. On the basis of the VSEPR theory, we expect a tetrahedral geometry.
However, the ion is bent due to the effect of the lone pairs on chlorine which is the central atom in the ion. This distorts the bond angle of the ion hence the ion has a bent molecular geometry.
At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.
The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.
At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.
At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults
<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction
<u>Explanation:</u>
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.
For the given chemical reaction:
<u>On the reactant side:</u>
Oxidation number of H = +1
Oxidation number of N = +5
Oxidation number of O = -2
Oxidation number of S = 0
<u>On the product side:</u>
Oxidation number of H = +1
Oxidation number of N = +2
Oxidation number of O = -2
Oxidation number of S = +6
As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.
The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.
Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction