When it says Carbon has 2 unpaired electrons, is referring to the fact that Carbon has an electron configuration of [He]2s22p2. Therefore Carbon has 2 unpaired electrons in the p sublevel.
Answer:
When the partially positive end of one water molecule is attracted to the partially negative end of another water molecule
Explanation:
<span>• Benzene has the formula of C6H<span>6 </span>whereas cyclohexane has the formula of C6H<span>12.
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Benzene is an aromatic compound but cyclohexane is not aromatic.
Benzene is an unsaturated molecule, but cyclohexane is saturated.
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Carbon atoms in the benzene ring have sp2 hybridization where carbon atoms in the cyclohexane have sp3 hybridization.</span>
Benzene has a planar structure whereas cyclohexane has chair conformations.<span>
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Explanation:
The shapes and relative energies of the orbitals s,p,d and f orbitals are given by the principal quantum number and the azimuthal quantum number.
The principal quantum number gives the main energy level and the azimuthal quantum number denotes the shape of the orbitals.
- For the principal quantum number, they represent the energy levels in which the orbital is located or the average distance of the orbital from the nucleus. It takes the number n = 1,2,3,4,5,6,7......
- The azimuthal quantum number(L) shows the shape of the orbitals in subshells accommodating electrons. The number of possible shapes is limited by the the principal quantum number.
L Name of orbital shape of orbital
0 s spherical
1 p dumb-bell
2 d double dumb-bell
3 f complex
Principal Azimuthal Orbital
Quantum Quantum Designation of
Number (N) Number(l) Sublevel
1 0 1s
2 0 2s
1 2p
3 0 3s
1 3p
2 3d
4 0 4s
1 4p
2 4d
3 4f
Learn more:
Atomic orbitals brainly.com/question/9514863
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Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2