Question

When calcium reacts with tellurium to form an ionic compound, each metal atom loses electron(s) and each nonmetal atom gains electron(s). There must be calcium atom(s) for every tellurium atom(s) in the reaction.

Answer 1

Answer:

When calcium reacts with tellurium it will donates its two electrons and these two electron will be accepted by tellurium.

There must be 1 calcium atom for every 1 tellurium atom in the reaction.

Explanation:

Calcium has atomic number of 20. Its electronic configuration is given as:

$[Ca]=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$

Calcium atom will loose 2 elections to attain electronic configuration of noble gas.

$[Ca^{2+}]=1s^2 2s^2 2p^6 3s^2 3p^6 4s^0$

$Ca\rightarrow Ca^{2+}+2e^-$

Tellurium has atomic number of 52. Its electronic configuration is given as:

$[Te]=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4s^24p^6 4d^{10} 5s2 5p4$

Tellurium atom will gain 2 elections to attain electronic configuration of noble gas.

$[Te^{2-}]=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4s^24p^6 4d^{10} 5s2 5p6$

$Te+2e^-\rightarrow Te^{2-}$

So, when calcium reacts with tellurium it will donates its two electrons and these two electron will be accepted by tellurium.

$Ca^{2+}+Te^{2-}\rightarrow CaTe$

There must be 1 calcium atom for every 1 tellurium atom in the reaction.