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rosijanka [135]
3 years ago
14

What is the hybridization of bromine in BrO2-?

Chemistry
2 answers:
Mashutka [201]3 years ago
6 0

Answer : The hybridization of the bromine in BrO_2^{-} is, sp^3

Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the bromine in BrO_2^{-} molecule.

The given molecule is, BrO_2^{-}

\text{Number of electrons}=\frac{1}{2}\times [7+1]=4

The number of electron pair are 4 that means the hybridization will be sp^3 and the electronic geometry of the molecule will be tetrahedral.

But as there are 2 atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent or angular.

Hence, the hybridization of the bromine in BrO_2^{-} is, sp^3

kap26 [50]3 years ago
5 0
<span>The hybridization of bromine must be sp^3.</span>
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Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.
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A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
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Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
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Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
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Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
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The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

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Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
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The cell representation of above reaction is given by;
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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a
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The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound            Amount

     Ni                        12.7 g

   CO                        1.98 g

Ni(CO)_4                  0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of Ni(CO)_4 = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M

\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M

\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M

For the given chemical reaction:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

The expression of equilibrium constant for the reaction:

K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0
3 years ago
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