All categories

3 years ago

Give the number of lone pairs around the central atom and the geometry of the ion IBr2.

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

Option E!

Explanation:

If we were to draw the lewis dot structure for IBr2 -, we would first count the total number of valence electrons ( " available electrons " ). Iodine has 7 valence electrons, and so does Bromine, but as Bromine exists in 2, the total number of valence electrons would be demonstrated below;

$7 + 7 * ( 2 ) =\\7 + 14 + 1 =\\22 Electrons$

Don't forget the negative on the Bromine!

Now go through the procedure below;

1 ) Place Iodine in the middle and draw single bonds to each of the bromine.

2 ) Add three lone pairs on each of the Bromine's

3 ) Now we have 6 electrons left, if we were to exclude the electrons shared in the " single bonds. " This can be placed as three lone pairs on Iodine ( central atom )!

The molecular geometry can't be linear, as there are lone pairs on the atoms. This makes it bent.

You might be interested in

The development of a nuclear power plant at an ocean site is expected to produce an enormous amount of electric power for a larg

Oksanka [162]

Answer:

B. The effects of warmed water on aquatic life

Explanation:

quizizz

8 0

3 years ago

Read 2 more answers

Usually hard but brittle Ionic or covalent

aleksandrvk [35]

Answer:

Ionic

Explanation: Ionic compounds tend to be hard and brittle while covalent compounds tend to be softer and more flexible.

Hope this helped!

7 0

3 years ago

Read 2 more answers

Calculate the volume occupied by 272g

cupoosta [38]

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

Explanation:

Given:

mass of methane( $CH_{4}$ ) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of methane( $CH_{4}$ ) = 16.0 grams

We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

6 0

3 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago

Which represents the correct equilibrium constant expression for the reaction below? Cu(s)+2Ag+(aq)<->Cu2+(aq)+2Ag(s)

balu736 [363]

K(eq) = concentration of products/concentration of reactant = [Cu+2] / [Ag+]^2
Activity of pure solid and liquid is taken as 1.
Hence last option is correct.
Hope this helps, have a great day ahead!
I hope this helps you alot
:)
:)
:0

4 0

3 years ago

Read 2 more answers