What is the order in which electrons start filling the orbitals?

Chemistry

2 answers:

liberstina [14]2 years ago

7 0

The answer is actually A, 1s 2s 2p 3s 3p 4s 3d 4p.

Elodia [21]2 years ago

4 0

Answer:

The correct answer is option A, that is, 1s 2s 2p 3s 3p 4s 3d 4p.

Explanation:

In quantum chemistry and atomic physics, the distribution of electrons of a molecule or an atom in atomic or molecular orbitals is known as the electronic configuration. The electronic configurations illustrate each electron as moving autonomously in an orbital. Mathematically, the configurations are illustrated by configuration state functions or Slater determinants. The electron starts filling the orbitals in an order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 7s 5f 6d and 7p.

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Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$