how many electrons does silver have to give up in order to achieve a sido noble gas electron configuration

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Answer:

Silver has to give up one electron.

Explanation:

Silver is a transition element and has a partially filled 4d- orbital having 9 electrons and a 5s orbital having 2 electrons in it. In order to achieve stability, silver must have completely filled d-orbital having a maximum number of 10 electrons in it.

Therefore, one electron from 5s orbital jumps to 4d orbital to make it stable. Now, 5s orbital has only one electron, and it will be easy for silver to lose this electron to attain a stable electronic configuration.

You might be interested in

What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)

Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

$pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414$

8 0

11 months ago

starting with methane ,name all the steps and write the equations involved in tetrachlorination.Remember all arrows sharing move

anygoal [31]

Answer:

See Explanation and image attached

Explanation:

Methane is an alkane. The commonest chemical reaction that alkanes undergo is substitution. During a substitution reaction, one or more atoms of hydrogen is/are replaced in the alkane.

In methane, in the presence of sunlight and molecular chlorine gas, a homolytic fission of Cl2 occurs to yield chlorine radicals in an initiation step.

The propagation steps involve reaction of the methane with chlorine radicals. Certain intermediates continue to be formed along the way until the tetrachlorination product is finally obtained.