Indique el valor de los números n, l,m y s para el último electrón de ₇N.(1 pt c/u) n = l = m = s =

n = 2

l = 1

m = 1

s = +1/2

Explanation:

₇N tiene la configuración electrónica;

1s2 2s2 2p3

Esto implica que este último electrón tiene los siguientes números cuánticos;

n = 2

l = 1

m = 1

s = +1/2

Este último electrón estará en un orbital de 2pz como lo muestran los números cuánticos enumerados anteriormente.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.