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Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

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

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options $XeF_{2}$ is the correct examples of linear molecules for five electron groups.

7 0

2 years ago

Carbon dioxide is a compound of the elements carbon and oxygen (CO2). What happened to the carbon and oxygen to create this comp

dexar [7]

Answer:

D. they exchanged electrons

6 0

1 year ago

What are standards or attributes of a design that can be measured.

ankoles [38]

Explanation:

attribute of a person that often cannot be measured directly but can be assessed using numbers of indicators or manifest variables

4 0

2 years ago

Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks

charle [14.2K]

Answer:

pH = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

4 0

2 years ago

Water contrasts on freezing ?

Airida [17]

No. When water first begins to cool down, it contracts. However, as it gets colder and eventually freezes, it begins to expand.

You can test this by freezing water in a water bottle: when you take it out of the freezer, the cap might have popped off or cracks may have formed in the sides of the bottle.

Answer: Water expands when frozen, not contracts.

8 0

3 years ago