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kolbaska11 [484]
3 years ago
15

What happens to the electronegativity as you go from Lithium to Francium? Why?

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
8 0

Answer:

Electronegativity in group 1 decreases as we go from Lithium to Francium.

Explanation:

Electronegativity is defined as the tendency of an element to attract an electron pair towards itself.

In a group generally this tendency decreases from top to bottom as the size of the atom increases and hence the positive nucleus get far from the outer orbital.

In the same way group 1 elements i.e. from Lithium to Francium electronegativity decreases.

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Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
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The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

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The slow step reaction is given as:

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The expression for the rate of reaction of slow reaction is:

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Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

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Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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