Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.
Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.
On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
PH measures the concentration of hydrogen ions on the log 10 scale. Thus, a pH 5 solution has 2 order of magnitude difference from that of pure water, which has a pH of 7.
Therefore,
. A solution of pH 5 has 100 times more hydrogen ions that that of pure water.
<span>In water, naoh almost completely separates into na+ and oh- ions. thus, naoh is Stong base.
Strong bases and strong acids completely ionized in the water,So when NaOH is added into water, it is completely ionized and forms Na+ and OH- ions and it shows that it is a strong base.</span>
