The electron configuration filling patterns of some elements in group 6b(6) and group 1b(11) reflect the increasing stability of half-filled and completely filled sublevels.
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What is electronic configuration?</h2>
The distribution of electrons in an element's atomic orbitals is described by the element's electron configuration. Atomic subshells that contain electrons are placed in a series, and the number of electrons that each one of them holds is indicated in superscript for all atomic electron configurations. For instance, sodium's electron configuration is 1s22s22p63s1.
Almost all of the elements write their electronic configurations in the same style. When the energies of two subshells differ, an electron from the lower energy subshell occasionally goes to the higher energy subshell.
This is due to two factors:
Symmetrical distribution: As is well known, stability is a result of symmetry. Because of the symmetrical distribution of electrons, orbitals where the sub-shell is exactly half-full or totally filled are more stable.
Energy exchange: The electrons in degenerate orbitals have a parallel spin and are prone to shifting positions. The energy released during this process is simply referred to as exchange energy. The greatest number of exchanges occurs when the orbitals are half- or fully-filled. Its stability is therefore at its highest.
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Answer:75%
Explanation:
First, the balanced reaction equation must be written out clearly as a guide to solving the problem. The molar masses of H3PO4 and K3PO4 are then calculated as they will be consistently required in solving the problem. The theoretical yield is obtained from the amount of H3PO4 reacted. Since 1 mole of H3PO4 yields 1 mole of K3PO4, 0.05 moles of H3PO4 yields 0.05 moles of K3PO4. The mass of K3PO4 is produced is then the product of 0.05 and it molar mass hence the theoretical yield. The % yield is calculated as shown.
<span>The two members of a conjugate acid-base pair differ by a proton.
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Answer:
grams O₂ = 134 grams
Explanation:
PV = nRT => n = PV/RT
P = 8.15atm
V = 12.2 Liters
R = 0.08206L·atm/mol·K
T = 16.0°C + 273 = 289K
n = (8.15atm)(12.2L)/(0.08206L·atm/mol·K)(289K) = 4.2 moles O₂
grams O₂ = 4.2 moles O₂ x 32g/mol = 134 grams O₂
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