Answer:
2. (C) K⁺; 3. (E) Hg⁺; 4. Hg⁺
Explanation:
We must first write the electron configurations of the different species.
(A) Fe²⁺
Fe: [Ar]4s²3d⁶
Fe²⁺: [Ar]3d⁶
When removing electrons from a transition metal ion, you remove the s electrons first.
(B) Cl
Cl: [Ne]3s²3p⁵
(C) K⁺
K: [Ar]4s
K⁺: [Ar]
(D) Cs
Cs: [Xe]6s
(E) Hg⁺
Hg: [Xe]6s²4f¹⁴5d¹⁰
Hg⁺: [Xe]6s4f¹⁴5d¹⁰
2. K⁺ has a noble gas configuration
3. Hg⁺ has electrons in f orbitals.
4. The electron configuration of Au is [Xe]6s4f¹⁴5d¹⁰, not [Xe]6s²4f¹⁴5d⁹, because a filled d subshell is more stable than a filled s subshell.
Thus, Hg⁺ is isoelectronic with Au.
False..........
it has so many consequences
From the calculations, the pH of the buffer is 3.1.
<h3>What is the pH of the buffer solution?</h3>
The Henderson-Hasselbach equation comes in handy when we deal with the pH of a buffer solution. From that equation;
pH = pKa + log[(salt/acid]
Amount of the salt = 25/1000 * 0.50 M = 0.0125 moles
Amount of the acid = 75/1000 * 1.00 M = 0.075 moles
Total volume = ( 25 + 75)/1000 = 0.1 L
Molarity of salt = 0.0125 moles/0.1 L = 0.125 M
Molarity of the acid = 0.075 moles/0.1 L = 0.75 M
Given that the pKa of lactic acid is 3.86
pH = 3.86 + log( 0.125/0.75)
pH = 3.1
Learn more about pH:brainly.com/question/5102027
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