Question

Формула высшего оксида элемента с электронной конфигурацией атома 1s22s22p63s23p64s23d104p1 – это 1) R2О; 2) R2О3; 3) R2О5; 4) R2О7.

Answer 1

Ответ:Сложим все электроны и получим их в количестве 31,в таблице Менделеева ищем элемент с зарядом +31 и это Ga. Высший оксид данного элемента- R203

Объяснение:Ответ под цифрой 2)

Answer 2

The formula of the highest oxide of the element with the electron configuration of the atom 1s22s22p63s23p64s23d104p1 is 2) R2O3. Here is a detailed explanation:

- The element with the electron configuration of 1s22s22p63s23p64s23d104p1 is gallium (Ga), which has 31 electrons and is in the 13th group of the periodic table. - The highest oxidation state of gallium is +3, which means it can lose three electrons from its outermost shell (4s23d104p1) and form a cation with the configuration of 1s22s22p63s23p64s23d10. - The most common oxide of oxygen is O2-, which has a charge of -2 and the configuration of 1s22s22p4. - To form a neutral compound, the charges of the cations and anions must be equal and opposite. Therefore, two gallium atoms (each with +3 charge) can combine with three oxygen atoms (each with -2 charge) to form gallium(III) oxide (Ga2O3). - The formula of gallium(III) oxide is R2O3, where R is the symbol of gallium. This is the highest oxide of gallium, as it cannot form a higher oxidation state than +3.